FORMULAE    AND    TABLES 

FOR   THE   CALCULATION    OF 

ALTERNATING  CURRENT   PROBLEMS 


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FORMULAE  AND  TABLES 

FOR  THE  CALCULATION  OF 

ALTERNATING  CURRENT 
PROBLEMS 


BY 

LOUIS   COHEN 

.'  I 


McGRAW-HILL  BOOK  COMPANY,  INC. 

239  WEST  39TH  STREET,  NEW  YORK 

6  BOUVERIE  STREET,  LONDON,  E.  C. 

1913 


COPYRIGHT,  1913,   BY  THE 
McGRAW-HILL  BOOK  COMPANY,  INC. 


Stanbopc 

P.    H.G1LSOK   COMPANY 
BOSTON,  U.S.A. 


PREFACE. 

THE  importance  of  mathematics  in  the  study  of  engineering 
problems  is  being  generally  recognized  now  and  fully  appreciated 
by  engineers.  The  rapid  development  in  alternating  current 
engineering  is  largely  due  to  the  skillful  utilization  of  mathe- 
matics in  the  study  of  the  many  complex  phenomena  which  arise 
in  connection  with  this  subject.  As  a  consequence  the  literature 
on  alternating  currents  is  distinctly  mathematical.  To  be  able 
to  utilize  fully  all  the  acquired  knowledge  on  this  subject,  one 
must  have  a  sufficient  mathematical  training  to  enable  him  to 
follow  the  derivations  of  the  many  formulae,  and  be  able  to  inter- 
pret them  correctly.  But  even  the  engineer  who  has  the  mathe- 
matical equipment  for  this  work  is  often  obliged  to  consult  a  good 
many  books  and  journals  before  he  finds  all  of  the  formulae  needed 
for  his  particular  problem. 

An  attempt  has  been  made  here  to  bring  together  in  a  small 
volume  all  the  important  formulae  which  are  necessary  and  help- 
ful in  the  solution  of  alternating  current  problems.  The  aim 
has  been  throughout  to  put  the  formulae  in  such  form  that  they 
can  be  immediately  applied  for  numerical  calculations;  and  in 
those  cases  where  it  seemed  desirable,  the  formulae  are  illustrated 
by  numerical  examples. 

In  a  work  of  this  kind,  which  is  in  the  nature  of  a  compilation, 
the  author  has  necessarily  drawn  freely  from  all  sources  of  infor- 
mation, but  in  most  cases  the  formulae  were  gone  over  carefully 
so  as  to  make  certain  that  they  are  free  from  errors.  Frequently 
the  formulae  were  derived  independently,  and  in  all  cases  an 
effort  was  made  to  put  them  in  the  simplest  possible  form.  A 
few  of  the  formulae  appear  here  for  the  first  time. 

The  author  is  fully  aware  that  accuracy  and  completeness 
are  of  first  importance  in  a  bpok  of  this  character.  To  this  end 
not  only  were  formulae  checked  with  care,  but  references  to 
original  authorities  are  generally  given.  The  author  will  appreci- 
ate a  notice  of  errors  discovered  in  the  work,  and  suggestions 


271056 


VI  PREFACE 

for  additions  of  other  material  which  might  increase  the  useful- 
ness of  the  book.  Both  the  scope  and  arrangement  of  the  present 
work  seem  logical  to  the  author,  although  a'  great  deal  of  material 
could  have  been  added,  but  not  without  the  risk  of  making  the 
book  "too  cumbersome  and  expensive  for  general  use. 

Many  thanks  are  due  to  Prof.  H.  H.  Norris  of  Cornell  Univer- 
sity for  his  careful  reading  of  the  manuscript  and  the  many  valu- 
able suggestions  he  has  made.  I  am  also  indebted  to  my  friend 
Dr.  P.  G.  Agnew  of  the  Bureau  of  Standards  for  going  over  care- 
fully parts  of  the  manuscript  and  checking  some  of  the  numerical 
examples.  The  diagrams  were  prepared  by  Mr.  B.  Holland  and 
Mr.  A.  F.  Van  Dyck  to  whom  I  wish  to  express  my  thanks. 

I  wish  also  to  express  my  appreciation  for  the  valuable  assist- 
ance rendered  by  the  publishers  and  their  technical  staff  in  the 
editing  of  this  book. 

LOUIS  COHEN. 

WASHINGTON,  D.  C,, 
September,  1913. 


CONTENTS. 

CHAPTER  I. 

PAGE 

RESISTANCE  AND  EDDY  CURRENT  LOSSES  IN  METALLIC  CONDUCTORS.       1 

Alternating  current  resistance  of  straight  cylindrical  conductors; 
series  formula;  high  frequency  formula;  general  formula  in  terms 
of  her  and  bei  functions  —  Current  penetration  in  cylindrical 
conductors  —  Resistance  of  looped  conductor  when  the  to  and 
fro  conductors  are  close  to  each  other  —  Resistance  of  concentric 
conductors;  formulae  for  low-frequency  currents,  high-frequency 
currents,  and  very  high-frequency  currents  —  Resistance  of  con- 
centric conductors  with  hollow  inner  conductor  —  Flat  conductors; 
alternating  current  distribution;  alternating  current  resistance; 
depth  of  current  penetration  —  Resistance  of  single  layer  coils 
for  alternating  currents;  general  formula;  approximate  formula 
suitable  for  high  frequencies;  approximate  formula  suitable  for  low 
frequencies  —  Alternating  current  resistance  of  large  slot  wound 
conductors  —  Distribution  of  induction  and  eddy  currents  in  a 
long  iron  pipe,  enclosing  an  alternating  current  —  Effect  of  grounded 
lead  covering  on  the  resistance  and  reactance  of  single  conductor 
cable  —  Effect  of  an  ungrounded  lead  covering  on  the  resistance 
and  reactance  of  a  single  conductor  cable  —  Leakage  conductance; 
between  a  cylindrical  conductor  and  an  infinite  plane;  between  two 
parallel  conductors;  three  core  cable —  Magnetic  flux  distribution 
in  iron  plates;  flux  density;  total  flux;  equivalent  depth  of  pene- 
tration; energy  loss  —  Magnetic  flux  distribution  in  iron  cylinders; 
flux  density;  total  flux;  eddy  currents;  energy  loss;  equivalent 
resistance  and  inductance  —  Magnetic  flux  distribution  in  iron  core 
of  toroidal  coil;  total  flux;  effective  resistance  —  Table  to  facilitate 
the  computation  of  the  resistance  of  cylindrical  conductors,  and  the 
magnetic  flux  distribution  in  cylindrical  conductors  —  Table  of 
specific  resistance  of  metallic  wires. 


CHAPTER  II. 
INDUCTANCE 48 

Mutual  inductance  of  two  coaxial  circles  —  Attraction  between 
circular  currents  —  Mutual  inductance  of  two  coaxial  coils  — 
Mutual  inductance  of  concentric  coaxial  solenoids  —  Self  inductance 

vii 


Viil  CONTENTS 

PAGE 

of  a  single  layer  coil  or  solenoid  —  Correction  factor  for  thickness 
of  insulation  on  wires  —  The  self  inductance  of  a  circular  coil  of 
rectangular  section  —  Self  inductance  of  circular  ring  —  Empirical 
formula  for  the  self  inductance  of  coils  —  Self  inductance  of  a  rect- 
angle —  Mutual  inductance  of  two  equal  parallel  rectangles  — 
Self  inductance  of  toroidal  coil  —  Inductance  of  linear  conductors 

—  Effect  of  frequency  on  the  inductance  of  linear  conductors  — 
Inductance  of  three-phase  transmission  lines  —  Mutual  inductance 
of  two  metallic  circuits  suspended  on  the  same  pole  —  Inductance 
of  grounded  conductors  —  Inductance  of  split  conductors;    con- 
ductor split  into  two  parts;  conductor  split  into  three  parts  —  Self 
inductance  of  two  parallel   hollow   cylindrical  conductors  —  Self 
inductance  of  concentric  conductors;  triple  concentric  conductors  — 
Self  inductance  of  a  circuit  formed  by  three  conductors  whose  axes 
lie  along  the  edges  of  an  equilateral  prism  —  Compound  conductors 

—  Series  or  parallel  arrangement  of  inductance  —  Inductance  of 
wires  in  parallel  —  Table  for  calculating  attractive  force  between 
two  circular  currents  —  Table  of  values  of  the  end  correction  in  the 
calculation  of  the  self  inductance  of  solenoids  —  Table  of  shape 
factors  for  certain  coil  proportions,  to  be  used  in  the  empirical  for- 
mula —  Table  of  inductances  of  complete  metallic  circuits. 


CHAPTER  III. 
CAPACITY 86 

Ratio  of  the  electrostatic  to  the  electromagnetic  unit  of  capacity 

—  Practical  unit  of  capacity  —  Parallel  or  series  arrangement  of 
condensers  —  Capacity  of  parallel  plates;  circular  plates;  correction 
for  edge  effects  —  Capacity  multiple  plate  condenser  —  Capacity 
of  concentric  spheres;   centers  of  spheres  separated  by  a  small  dis- 
tance —  Capacity  of  disk  insulated  in  free  space  —  Capacity  coeffi- 
cients of  two  spheres  —  The  laws  of  attraction  and  repulsion  between 
spheres  when  the  potentials  are  given  —  Capacity  of  linear  con- 
ductors;  one  overhead  wire,  the  ground  as  the  return  conductors; 
two  overhead  wires;  three  overhead  wires;  four  overhead  wires  — 
Capacity  of  condenser  formed  by  two  long  parallel  cylinders  of  equal 
radii;    unequal   radii;    influence  of  the  earth  as  a  nearby   con- 
ducting plane  —  Capacity  two  metallic  circuits  suspended  on  the 
same  pole  —  Capacity  three-phase  transmission  lines  —  Capacity 
of  horizontal  antennae  —  Capacity  concentric  cylinders;    axes  of 
cylinders   displaced  with  respect    to    each   other  —  Capacities   of 
cables;  concentric  cable;  triple  concentric  cable  —  Capacity  of  two 
core  cable;    effective  capacity  under  different  conditions  —  Capa- 
city of  three  core  cables;  effective  capacity  under  various  conditions 

—  List  of  capacities  that  can  be  obtained  from  a  three  core  cable 

—  Table  of  specific  inductive  capacities  of  solids  —  Table  of  specific 
inductive  capacities  of  liquids. 


CONTENTS  ix 

CHAPTER  IV. 

PAGE 

ALTERNATING  CURRENT  CIRCUITS 126 

Complex  waves  —  Characteristic  features  of  different  forms  of 
alternating  current  and  pressure  curves  —  Effective  value  of  com- 
plex wave  —  Form  factor  —  Curve  factor  —  Current  and  phase 
angle  in  circuit  of  inductance  and  resistance  —  Current  and  phase  . 
angle  in  circuit  containing  capacity  and  resistance;  voltage  across 
condenser  —  Circuits  containing  inductance  and  capacity;  res- 
onance condition;  voltage  across  condenser  —  Impedances  in 
series  —  Parallel  circuits  each  having  inductances  and  resistance  — 
A  system  of  branched  circuits  in  parallel  each  having  inductance  and 
resistance  —  Inductance  and  resistance  shunted  by  condenser; 
resonance  condition  —  Inductive  load  shunted  by  condenser  — 
Two  branch  circuits  in  parallel,  one  having  inductance  and  resistance, 
and  the  other  inductance  capacity  and  resistance  —  The  branched 
circuits  as  well  as  the  main  circuits  having  inductance  capacity 
and  resistance  —  Mutual  inductance  between  the  branched  circuits 

—  Power  factor;     simple  wave;    complex  wave  —  Power  trans- 
mission;  maximum  power;   efficiency;   ratio  of  e.  m.  f.  at  receiver 
and  generator  —  Power  transmission   inductive  load;    maximum 
power;    efficiency;    ratio  of  e.  m.  f.  at  receiver  and  generator  — 
Power  transmission  calculations;    assuming  capacity  of  the  line 
shunted  across  the  middle  of  the  line;    half  of  the  line  capacity 
shunted   across   each    end   of  the  line  —  Air  core  transformer  — 
Resonance  transformer;    condition  for  maximum  current  in  the 
secondary  circuit  —  Transformer  having  a  condenser  only  in  the 
secondary  circuit  —  Transformer  having  condenser  in  the  primary 
circuit  —  Iron  core  transformer;  general  design  formula;  efficiency; 
ratio  of  terminal  voltages  for  inductive  and  non-inductive  load; 
regulation;  phase  angles  —  Current  transformer;  ratio  of  transfor- 
mation; phase  angle. 

CHAPTER  V. 
TRANSIENT  PHENOMENA 171 

Rise  and  decay  of  currents  in  an  inductive  circuit  on  closing 
and  opening  the  circuit  —  Rise  and  decay  of  current  on  closing 
and  opening  on  alternating  inductive  circuit  —  Current  and  volt- 
age on  condenser  on  closing  a  direct  current  circuit  containing 
resistance  and  capacity,  no  inductance  —  Circuit  containing  re- 
sistance inductance  and  capacity;  current  and  voltage  on  charging 
and  discharging  for  logarithmic  case;  critical  case;  trigonometric 
case  —  Frequency  of  oscillations;  damping;  logarithmic  decrement 

—  Effective  current  —  Electrostatic  energy  —  Alternating  current 
circuits  containing  resistance,  inductance  and  capacity;  transient 
term  on  closing  and  opening  circuit;  logarithmic  case;  critical  case; 
trigonometric  case  —  Divided  circuits;    transient  terms  for  change 


X  CONTENTS 

PAGE 

in  resistance  in  either  branch  —  Transient  currents  in  transmission 
lines;  load  short  circuited;  generator  short  circuited  —  Mutual  in- 
ductance; values  of  transient  terms  which  arise  on  sudden  change 
in  resistance  in  primary  or  secondary  circuit;  current  rise  in  circuits 
when  an  e.  m.  f .  is  introduced  in  one  circuit  —  Oscillation  trans- 
former having  inductance  and  capacity  in  series;  currents  and 
voltages  in  the  circuits;  frequencies;  ratio  of  voltages  —  Direct 
coupling;  syntonized  circuits;  frequencies. 

CHAPTER  VI. 
DISTRIBUTED  INDUCTANCE  AND  CAPACITY 200 

The  general  equations  for  current  propagation  along  wires  — 
Propagation  constant  of  the  line;  attenuation  constant;  velocity 
constant  —  Approximate  values  of  the  propagation  constants; 
the  inductance  is  large;  the  inductance  negligibly  small  —  Prop- 
agation constants  expressed  as  a  function  of  the  ratio  of  inter- 
axial  distance  and  radius  of  wire  —  General  solution  for  current  and 
voltage  on  line  expressed  in  complex  quantities  and  hyperbolic 
functions  —  Infinite  line;  current  and  voltage  —  Line  of  finite 
length  and  open  at  receiving  end — Short  lines  —  Line  grounded 
at  receiving  end  —  Line  short  circuited  at  receiving  end  —  Power 
transmission;  voltage  and  current  given  at  the  receiving  end; 
voltage  and  current  given  at  the  generator  —  Hyperbolic  formulae 

—  Approximate    formulae    for    short    lines  —  Generator    voltage 
and  impedance  at  receiving  end  known  —  Transient  phenomena 
in  circuits  having  distributed  resistance  and  capacity;    cable  open 
at  one  end  and  a  constant  continuous  e.m.f.  suddenly  applied 
at  the  other  end  —  A  charged  cable  having  a  constant  e.m.f.  applied 
between  one  end  of  the  cable  and  ground  is  suddenly  grounded 
at    the    other    end  —  Circuits    containing    distributed    resistance, 
inductance  and  capacity;  line  oscillations  —  Line  is  grounded  at  one 
end  and  open  at  the  other  end  —  Line  grounded  at  both  ends  — 
Line  open  at  both  ends  —  Line  closed  upon  itself  —  Table  of  hyper- 
bolic functions  of  complex  quantities. 

CHAPTER  VII. 
MATHEMATICAL  FORMULAE 264 

Exponential  and  logarithmic  formulae  —  Trigonometric  formulae 

—  Hyperbolic  formulae  —  Complex  quantities  —  Miscellaneous  for- 
mulae —  Miscellaneous  functions  —  Interpolation  formula  —  Table 
of  Naperian   logarithms  —  Table  of  exponential  and  hyperbolic 
functions  —  Table  of  trigonometric  functions  —  Table  of  binomial 
coefficients  for  interpolation  by  differences. 

INDEX..  277 


ALTERNATING  CURRENT  PROBLEMS 


CHAPTER  I 

RESISTANCE   AND    EDDY-CURRENT   LOSSES   IN 
METALLIC    CONDUCTORS 

THE  resistance  of  metallic  conductors  is  higher  for  alternating 
currents  than  for  continuous  currents  due  to  the  so-called  skin 
effect.  In  the  case  of  alternating  currents  the  distribution  of  the 
current  is  not  uniform  throughout  the  entire  cross  section  of  the 
conductor,  but  tends  to  concentrate  more  towards  the  surface  of 
the  conductor,  and  as  a  consequence  there  is  a  corresponding  in- 
crease in  resistance.  The  increase  in  resistance  for  alternating 
currents  depends  on  the  conductivity  and  permeability  of  the 
conductor  and  on  the  frequency.  The  higher  the  conductivity, 
permeability  and  frequency  the  greater  the  increase  in  resistance. 

Besides  the  PR  loss,  there  is  another  source  of  energy  loss  in 
alternating  current  circuits  which  does  not  occur  in  continuous 
current  circuits.  This  is  the  eddy-current  loss  in  outside  con- 
ductors. When  an  alternating  current  passes  through  a  con- 
ductor it  creates  an  alternating  magnetic  field  which  will  induce 
currents  in  conductors  placed  in  that  magnetic  field  and  thus 
causes  loss  of  energy.  This  is  particularly  marked  in  the  case  of 
iron  bodies  placed  in  a  strong  alternating  magnetic  field  as  in 
the  case,  for  instance,  of  a  cylindrical  iron  core  inserted  within  a 
cylindrical  coil.  This  loss  of  energy  due  to  eddy  currents  in  out- 
side conducting  bodies  may  be  looked  upon  as  an  equivalent  in- 
crease in  resistance  of  the  circuit.  In  fact  the  outside  conductor 
acts  like  a  closed  secondary  of  a  transformer  which  absorbs  energy 
from  the  primary. 

In  this  chapter  on  resistance  formulae  we  shall  consider  not 
only  the  resistance  of  conductors  proper,  but  also  the  increase 
in  effective  resistance  caused  by  eddy-current  losses  in  metallic 
bodies  which  are  placed  in  magnetic  fields. 

The  general  problem  of  determining  the  distribution  of  alter- 
nating currents  in  conductors  of  any  shape  offers  considerable 

1 


FORMULAE  AND  TABLES  FOR  THE 

mathematical  difficulties,  and  has  only  been  worked  out  for  a 
limited  number  of  special  cases.  We  shall  endeavor  to  give  here 
all  the  important  available  formulae  with  numerical  examples  to 
illustrate  their  application. 

Straight  Cylindrical  Conductors.  —  The  problem  of  determin- 
ing the  effective  resistance  of  wires  for  alternating  currents  has 
engaged  the  attention  of  many  able  physicists  and  several  for- 
mulae have  been  deduced  for  this  case.  Owing  to  the  importance 
of  this  problem  in  engineering  practice,  we  shall  give  here  all  the 
formulae  available,  with  a  brief  discussion  of  their  limitation  and 
application. 

In  what  follows  let 

R  =  resistance  for  alternating  currents. 
RO=  resistance  for  continuous  currents, 
co  =  2  TT  X  frequency  =  2  irn. 
I  =  length  of  wire  in  cm. 
ju  =  permeability. 

p  =  specific  resistance  in  C.G.S.  units. 
d  =  diameter  of  wire  in  cm. 
The  permeability  JJL  is  assumed  to  be  constant. 
The  form  in  which  the  formula  for  the  resistance  of  cylindrical 
conductors  for  alternating  currents  was  first  given  is  as  follows:* 


For  non-magnetic  materials 

1    a,4*4 


(1) 


(2) 


.  ****<*  V  I     .  .  .  (3) 

~  "" 


Formula  (1)  may  also  be  written  in  the  following  form: 


Formulae  (1)  to  (3)  are  applicable  only  for  low  frequencies. 

When  the  diameter  of  the  wire  is  not  very  small  and  the  fre- 
quency fairly  high,  the  series  in  the  above  formulae  is  very  slowly 
convergent,  and  it  will  require  more  terms  in  the  series  for  an 
accurate  determination  of  the  resistance. 

*  See  J.  A.  Fleming,  "The  Principles  of  Electric  Wave  Telegraphy  and 
Telephony,"  second  edition,  p.  117. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    3 

Example.  —  Copper  wire  No.  0  B.  &  S. 

d  =  0.8252  cm.,     p  =  1700,     n  =  500. 
R  =  Ro  }1  +  0.081  -  0.0055  +  -.  .  .  |  .  1.076  RQ. 

This  is  a  rapidly  convergent  series.  The  increase  in  resistance 
for  this  case  is  about  7.6  per  cent. 

If,  however,  in  the  same  example  we  put  n  =  1500  we  get 

R  =  Ro\l+  0.729  -  0.445  +•••}• 

The  series  is  very  slowly  convergent;  it  will  require  therefore 
many  more  terms  to  determine  the  value  of  R  accurately.  For 
still  higher  frequencies  or  larger  diameter  of  wire,  the  series  in 
formulae  (1)  to  (3)  will  be  still  more  slowly  convergent,  and  there- 
fore the  formulae  will  not  be  suitable.  In  the  case  of  magnetic 
material,  as  that  of  iron  wire,  the  formulae  is  only  applicable  for 
very  low  frequencies. 

When  the  frequency  is  very  high,  we  have  the  following  con- 
venient formulas,* 

R  = 

R  = 


8  p 

For  copper  wire,  /*  =  1,  p  =  1700  at  ordinary  temperature,  and 
formula  (4)  reduces  to 


Formulae  (4)  and  (5)  are  applicable  only  when  —  —  -  has  a  value 

P 
much  greater  than  unity,  of  the  order  10  or  more. 

It  is  obvious  that  the  above  formulae  (1)  to  (5)  have  only  a 
limited  application.     Formulae  (1)  to  (3)  can  be  used  only  when 

—  is  small,  and  formulae  (4)  and  (5)  when  --  -  is  large.     Physi- 

cally it  means  that  the  first  set  of  formulae  are  applicable  to  cases 
in  which  current  traverses  almost  the  entire  cross  section  of  the 
conductor,  while  formulae  (4)  and  (5)  are  applicable  to  cases  in 
which  the  current  is  confined  mostly  to  a  thin  surface  layer  of  the 
conductor. 

*  Lord  Rayleigh,  "The  Self-induction  and  Resistance  of  Straight  Con- 
ductors."     Philosophical  Magazine,  Ser.  V,  May,  1886,  Vol.  21,  p.  381. 


4  FORMULAE  AND  TABLES  FOR  THE 

Recently  Professor  Pedersen*  extended  the  work  on  this  prob- 
lem so  as  to  cover  all  cases  of  current  distribution;  he  also  worked 
out  a  valuable  set  of  tables  which  are  reproduced  here,  see  Table  II, 
p.  43.  The  formula  is  as  follows: 

R  =  R0r  (ma),  (6) 

.      ^  _  ma  ber  (ma)  bei'  (ma)  —  bei  (ma)  ber'  (ma)          ,  . 
'^~  ber"  (ma)  +  bei'2  (ma) 


where 

p 

and  a  =  radius  of  wire. 

The  value  of  r  (ma)  for  ma  from  0  to  6  is  given  in  Table  II  and 
for  values  of  ma  larger  than  6,  the  expression  for  r  (ma)  as  given  in 
(7)  reduces  to 


ft 
r  (ma)  =  0.35355  ma  +  0.25  +  —    -•  (8) 

ma 

In  Table  I,  p.  42,  are  given  the  values  of  m  for  copper  wire  for 
different  frequencies.  By  the  aid  of  Tables  I  and  II  and  formulae 
(6)  and  (8)  we  can  compute  quite  readily  the  alternating  current 
resistance  of  wires  of  any  diameter  and  all  frequencies. 

To  further  facilitate  the  work  of  computing  the  resistance  of 
copper  wires  there  are  given  in  Figs.  1  and  2  curves  expressing  the 

TT> 

value  77-  as  a  function  of  Van  for  values  of  Van  from  1  to  1000. 
/to 

Example.  —  Copper  wire  No.  0  B.  &  S., 

d  =  0.8252  cm.          p  =  1700,        n  =  500, 
m  =  4.824,  ma  =  1.99. 

From  Table  II  we  get  r  (ma)  =  1.077, 
hence  R  =  1.077  R0, 

which  agrees  with  result  obtained  for  the  same  example  by  formula 
(3). 

If  we  put  in  the  above  example  n  =  1500  we  have 
m  =  8.346,        ma  =  3.443. 

*  P.  O.  Pedersen,  Jahrbuch  der  DrahOosen  Telegraphic  und  Telephonie,  Vol. 
4,  pp.  501-515. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    5 


4.0 
3.5 
3.0 

2.5 
< 
2.0 

1.5 

1.0 

y 

x 

x 

X 

4 

X 

x 

X 

X 

X 

/ 

% 

X 

X 

Curve  showing  the  change  in  resistance 
of  copper  wire  for  alternating  currents 

R  =  Resistance  at  frequency  (n) 
n  =  Frequency  (  cycles  per  sec.  ) 
RO=°  Resistance  for  direct  current 

a  =>  Radius  of  wire  in  centimeters 

<x 

X 

— 

X 

? 

• 

«s 

10 


20     25 
i 
FIG.  1. 


30 


35     40 


46    80 


8U 
75 
70 
65 
60 
55 
50 

4 

•< 

40 
35 
30 
25 
20 
15 
10 
5 

/ 

7 

X 

X 

x 

X 

x 

r 

X 

/ 

x 

X 

* 

x 

~> 

/ 

Curve  showing  the  change  in  resistance 
of  copper  wire  for  alternating  currents 

R=  KR0 
R  =  Resistance  at  frequency  (  n  ) 
n  =  Frequency  (  cycles  per  sec.  ) 
RQ=  Resistance  for  direct  current 

Q>  =  Radius  of  wire  in  centimeters 

x 

' 

x 

* 

X 

X 

/ 

100 


200 


400    500    600 

a 
FIG.  2. 


700 


900   1000 


6  FORMULAE  AND  TABLES  FOR  THE 

From  Table  II,  r  (ma)  =  1.470, 

hence  R  =  1A7  RQ. 

For  a  frequency  100,000, 

m  =  68.22,        ma  =  28.15. 

Computing  r  (ma)  by  formula  (8)  we  get 

r  (ma)  =  10.2, 
hence  R  =  10.2  R0. 

Current  Penetration  in  Cylindrical  Conductors.  —  It  was 
pointed  out  above  that  the  increase  in  resistance  for  alternating 
currents  is  due  to  the  fact  that  the  current  is  not  uniformly  dis- 
tributed over  the  entire  area  of  the  conductor.  The  intensity  of 
the  current  diminishes  from  the  surface  to  the  center  of  conductor. 
The  amplitude  of  the  current  at  any  point  distance  r  from  the  axis 
is  given  by  the  formula  * 


where    IQ  =  current  density  on  surface  of  conductor. 
a  =  radius  of  conductor  in  cm. 
e  =  base  of  (Naperian)  logarithms. 

There  is  also  a  change  in  phase  in  the  current  from  point  to 
point  along  the  radial  axis.  The  intensity  of  the  current  dimin- 
ishes in  a  geometric  progression  as  the  distance  from  the  sur- 
face increases  in  arithmetical  progression.  For  a  distance  from 


the  surface  a  —  r  =  V       —  »  the  density  of  the  current  will  be 


reduced  to  -  =  s-==  its  surface  value.     V       —  is  taken  as  the 

€          2.77  V    2  7T/ICO 

measure  of  the  thickness  of  current  penetration,  that  is,  we  may 
consider  the  current  concentrated  in  a  cylindrical  shell  of  thick- 

ness \~^  —  )    and  the  sectional  area  of  the  shell  as  the  effective 


area  for  the  passage  of  the  current.     Denoting  by  5  the  depth 
of  current  penetration,  that  is,  the  distance  from  the  surface  at 

*  See  J.  J.  Thomson,  "  Recent  Researches  in  Electricity  and  Magnetism,"  p.  281. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    7 
which  the  current  density  is  -  part  of  the  density  at  the  surface,* 


We  have  for  copper  wire  (p  =  1700) 


for  aluminum  wire  (p  =  2800), 

«  =  M  .      \ 

and  for  iron  wire  (p  =  10,000,  n  =  1000) 

5=S" 

Example.  —  Let  n  =  500. 
For  copper  wire, 

5=-^  =  0.29 

Vsoo 

that  is  the  current  penetrates  to  0.29  times  the  radius  below  the 
surface  of  the  conductor.  Assuming  the  radius  of  the  conductor 
to  be  1  cm.,  the  total  area  of  the  conductor  is  TT  sq.  cms.  The  area 
through  which  the  alternating  current  is  passing  is  TT  (I2  —  0.7 12)  = 
0.496  TT  sq.  cm.  This  is  about  one-half  the  actual  area  and  there- 
fore the  effective  resistance  is  about  double. 

By  formula  (6),  we  have  for  this  case,  m  =  4.824,  ma  =  4.824, 
and  r  (ma)  =  1.98. 

The  alternating  current  resistance  is  1.98  times  the  continuous 
current  resistance,  which  agrees  closely  with  the  value  obtained 
above. 

Return  Conductor.  —  When  conductors  are  placed  close  to 
each  other,  as  in  the  case  of  cables,  the  current  distribution  is 
effected  by  the  presence  of  the  return  conductor.  The  greatest 
current  density  occurs  on  the  sides  of  the  conductors  which  face 
each  other.  For  two  conductor  cables  and  low  frequencies  up  to 

*  See  "  Theorie  der  Wechselstrome  "  von  J.  L.  LaCour  und  O.  S.  Bragstad, 
Zweite  Auflage,  p.  568. 


8  FORMULAE  AND  TABLES  FOR  THE 

about  2000  cycles  per  second  we  have  the  following  formula  for 
the  alternating  current  resistance.* 


where  d  =  diameter  of  wire 

and  h  =  interaxial  distance  between  wires. 

For  copper  cables  formula  (11)  may  be  put  in  the  following  more 
convenient  form 


For  aluminum  cables 


For  magnetic  materials,  the  distance  between  the  wires  has 
very  little  influence  on  the  current  distribution,  and  we  may 
therefore  use  the  same  formula  as  that  for  a  single  conductor. 
Example.  —  Two-conductor  copper  cable  of  No.  1  B.  &  S.  wires. 
Let  n  =  300,  d  =  0.73  cm.,  h  =  1  cm. 

R=R0\l  +  [0.70  +  0.40]  0.0256  -  [0.4  +  4.256]  0.00065} 
=  #o  \  1  +  0.028  -  0.0030|  =  1.025  RQ, 

an  increase  in  resistance  of  about  2.5  per  cent. 

For  a  single  conductor  of  the  same  diameter  and  the  same 
frequency  we  have 

R  =  R0  (1  +  0.018  -    •  •  •  )  =  1.018  Bo, 

an  increase  of  only  1.8  per  cent,  hence  the  presence  of  the  other 
conductor  introduced  another  increase  in  resistance  of  0.7  per  cent. 
For  very  high  frequencies  the  resistance  of  a  two-conductor 
cable  is  given  by  the  expression: 


2    «- 


Formula   (12)  is  applicable  only  when  —  -  iV-  is  small  com- 

7T  d    »    H 

pared  with  unity. 

*  G.  Mie,  Wied.  Ann.,  1900.     See  also  "Theorie  der  Wechselstrome,"  von 
J.  L.  LaCour  und  O.  S.  Bragstad,  Zweite  Auflage,  p.  569. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    9 

Example.  — 

Let  d  =  0.5  cm.,  h  =  0.75  cm.,  p  =  1700  (copper),  n  =  106. 


t       =  0.05, 

Trd  V  n 

which  is  small  compared  with  unity,  and  hence  we  may  use  formula 
(12)  for  this  case. 

R  =  Ro  (51.1  +  0.125  -  0.335)  =  50.9  R0. 

The  resistance  is  about  fifty-one  times  the  resistance  for  con- 
tinuous currents. 

For  a  single  conductor  of  the  same  diameter  and  the  same  fre- 
quency, the  resistance  is  only  about  twenty  times  the  continuous 
current  resistance. 

Resistance   of  Concentric   Conductors.  —  This  problem  has 
been  investigated  in  a  very  able  manner  by  Dr.  Alexander  Russell,* 
who  derived  a  general  expression  for  the  resistance  and  inductance 
of  concentric  conductors  and  from  it  he  deduced  several  simple 
formulae  suitable  for  low  frequencies,  high  frequencies  and  very 
high  frequencies.     The  general  expression  is  somewhat  unwieldy 
and  not  convenient  for  numerical  calculations,  and  we  shall  there- 
fore give  here  only  the  simplified  formulae  for  particular  cases. 
Inner  Conductor  Solid  Metal  Cylinder. 
Let   a  =  radius  of  inner  conductor  in  cm. 

b  =  inner  radius  of  concentric  outer  cylinder  in  cm. 
c  =  outer  radius  of  concentric  outer  cylinder  in  cm. 
n  =  frequency. 
p  =  resistivity  in  C.G.S.  units. 


m  =  2ir 

P 

For  direct  currents  we  have 


The  first  term  in  the  right  hand  part  of  equation  (13)  is  the 
resistance  of  inner  conductor  and  the  second  term  the  resistance  of 
outer  conductor. 

*  Alexander  Russell,  "  The  Effective  Resistance  and  Inductance  of  a  Con- 
centric Main,  and  Methods  of  computing  the  Ber  and  Bei  and  Allied  Functions," 
Phil.  Mag.,  April,  1909. 


10  FORMULAE  AND  TABLES  FOR  THE 

For  low-frequency  currents  (ordinary  commercial  frequencies 
25  to  60  cycles) 


where 


Cm*a*\ 
1  +  1927  +  Ro11  +  ™*  (pl  +  p*  +  P3?)},       (14) 

(7c2-62)(c2-62)  62c4 

Pl  =  ~192~  p2  =  8(c2-&2)' 


4(0*  - 
For  high-frequency  currents,  when  ma  is  greater  than  5,  that  is 

>5    or 
p 

540 
and  for  copper,  n  >  —  £- 

the  following  formula  is  applicable: 
R==p™_  (1.1. 


8\/2ra2a2) 

pm      sinh  m  (c  -  6)  A/2  +  sin  m  (c  -  b)  V2 
2  V2irb  coshm  (c  -  6)  A/2  -  cosm  (c  -  b)  \/2 

The  first  term  gives  the  resistance  of  the  inner  conductor  and  the 
second  term  the  resistance  of  the  outer  conductor. 

When  the  frequency  is  very  high  and  c  is  not  nearly  equal  to  6, 
we  have  the  following  expression  for  the  resistance  of  a  concentric 
conductor  : 

(16) 


2      27T&2  \a 

Example.  —  Let  the  radius  of  inner  conductor,  a,  be  1  cm. 
and  the  external  and  internal  radii  of  outer  conductor  2  and 
1.73  cm.  respectively.  This  will  give  equal  sectional  areas  for 
the  two  conductors  which  is  usually  the  case  in  practice.  If 

we  assume  a  frequency  of  25,  we  get  m  =27r\/  —  =  2  ?r  V/  T^™ 

p  "    1  /  UU 

=  1.08,  and  for  this  value  of  m  with  the  radii  as  given  in  this 
example,  formula  (14)  is  applicable.  Introducing  these  constants 
in  (14), 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     11 


m4  =  1.36,  pi  =  0.1302,   P2  =  6,  P3  =  -  48,  {  =  loge  ^  =  0.1438 
and 

R  =  Ri 


T§) 


L36  x  a0004) 


=  #»  (1.007)  +  #o  (1.00054). 

The  resistance  of  inner  conductor  is  increased  by  0.7  per  cent 
while  the  resistance  of  outer  conductor  is  increased  only  by  0.05 
per  cent.  The  increase  in  resistance  of  inner  conductor  is  about 
fourteen  times  that  of  outer  conductor. 

If  in  the  above  example  we  make  n  =  1000,  m  =  2r  V —  =  7.4, 

p 

ma  is  greater  than  5  and  we  may  therefore  use  formula  (15) 
sinh  m  (c  -  b)  A/2  =  8.2,          cosh  m  (c  -  b)  \/2  =  8.3, 
sin  m  (c  -  6)  V2   =  0.332,       cos  m  (c  -  b)  \/2  =  -  0.943, 

8.2  +  0.332 


R  =  1987  (0.707  +  0.068  +  0.005)  +  818  X 
=  109  (1550  +  753). 


8.3  +  0.943 


For  direct  current  the  resistance  is  R  =  109  (541  +  541). 

The  resistance  of  inner  conductor  was  increased  to  about  three 
times  the  direct  current  resistance,  while  the  resistance  of  outer 
conductor  was  increased  only  to  one  and  five-tenths  tunes  the 
direct  current  resistance. 

Concentric  Conductors  with  Hollow  Inner  Conductor.  —  If 
we  denote  by  a\  and  a2  the  internal  and  external  radii  of  inner 
conductor,  and  as  before  c  and  6  denote  the  outer  and  inner  radii 
of  outer  conductor,  we  have  for  low-frequency  currents  correspond- 
ing to  formula  (14), 


7r(a2: 


+  w4 
02 


192 


a2 


7T  (C2  ~  62) 

6V 

8  (c2  -  62) 


6V 


(17) 


12 


FORMULAE  AND  TABLES  FOR  THE 


— x-*- 


For  high  frequencies  corresponding  to  formula  (16),  we  have 

„  pm      sinh  m  (a?  —  a\)  V2  +  sin  m  (a*  —  ai)  V2 

2  Trai  A/2  cosh  m(a2  —  ai)  \/2  —  cos  m(az  —  a\)  A/2 

\    (18) 
pm      sinh  m  (c  —  b)  v  2  +  sin  m  (c  —  b)  v  2 

2  7r6  V2  cosh  m(c  -  6)  V2  -  cos  m(c-b)  V2 

In  the  above  two  formulae,  (17)  and  (18),  Ri  is  the  resistance  of 
inner  conductor  and  RO  the  resistance  of'  outer  conductor. 

Flat   Conductors.  —  Resistance    and   distribution   of   current 
density  in  flat  conductors  with  alternating  currents.* 

Let  Fig.  3  represent  a  section  of 
the  conductor,  the  current  flowing 
in  the  direction  perpendicular  to  the 
plane  of  the  paper.  We  assume  that 
the  conductor  is  very  thin  compared 
with  its  width,  as  in  the  case  of  flat 

ribbon  conductors. 
v 

Let    2  a  =  thickness  of  conductor 

in  cm. 
p  —  specific    resistance    in 

C.G.S.  units. 
n  =  frequency. 
-^  p  =  permeability. 

FIG'3-  m=2*\J^. 

The  current  Ix  at  any  point  distant  x  from  the  center  is  given  by 
the  expression 

E  f  cosh  2  mx  +  cos  2  mx  j>?  ,^. 

x  ~  p  {  cosh  2  ma  +  cos  2  ma ) 

Formula  (19)  gives  only  the  amplitude  of  the  current  at  any 
point  in  the  conductor,  but  there  is  also  a  continuous  change  in 
phase  of  the  current  from  point  to  point  from  the  surface  to  the 
center  of  the  conductor. 

The  current  density  at  the  center  of  the  conductor,  x  =  0,  is 


(20) 


E 

p  (cosh  2  ma  +  cos  2  ma)2" 


*  See  Ch.  P.  Steinmetz,  "Theory  and  Calculation  of  Transient  Electric 
Phenomena  and  Oscillations,"  Chapter  VII. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     13 

The  mean  value  of  the  current  for  the  entire  section  of  the 
conductor  is 

E       \  cosh  2  ma  —  cos  2  ma  )?  ,~i  \ 

m      pma  V2  I  cosh  2  raa  +  cos  2  ma  ) 

If  we  denote  by  R0  the  direct  current  resistance  and  by  R  the 
effective  resistance  for  alternating  currents,  we  have 

R  /p.  (  cosh  2  ma  +  cos  2  ma  ^  £  /00, 

=  ma  V2    <  -  7-jr  -  !  -  jr  -  >     •  (22) 

.Ro  (  cosh  2  ma  —  cos  2  ma  ) 

Special  Cases.  —  If  2  ma  is  sufficiently  large  so  that  we  can  neglect 
cos  2  ma  as  compared  with  cosh  2  ma,  formula  (22)  reduces  to 


=  27T     2\a.  (23) 

o  >     p 

The  increase  in  resistance  is  directly  proportional  to  the  thickness 
of  plate,  the  square  root  of  frequency  and  permeability,  and 
inversely  as  the  square  root  of  the  resistivity. 

If  in  formula  (19)  2  ma  is  sufficiently  large  so  that  cos  2  ma  and 
e-2ma  can  be  neglected  compared  with  e2ma,  then  for  points  not 
far  below  the  surface  of  the  conductor,  formula  (19)  assumes  the 
form 

E  E 

Ix  =*£«-*<•-*>  =  -€-"»,  (24) 

P  P 

where  s  =  a  —  x  is  the  distance  below  the  surface  of  conductor. 
The  density  of  the  current  diminishes  in  geometric  progression  as 
the  distance  from  the  surface  increases  in  arithmetic  progression. 
For  the  value  of  ms  =  1,  the  current  density  is  reduced  to  about 
one-third  its  maximum  value,  and  from  that  point  on  the  current 
density  diminishes  with  great  rapidity;  hence  we  can  take  the  value, 

.     (25) 

as  the  measure  of  the  thickness  of  current  penetration,  tha*t  is,  we 
may  consider  the  current  flowing  through  only  two  surface  layers 
of  the  conductor  of  thickness  s  given  by  formula  (25). 

We  have  obtained  an  identical  expression  for  the  case  of  cylindri- 
cal conductors;  see  formula  (10).  It  is  obvious,  of  course,  that  in 
the  design  of  conductors  for  alternating  currents,  the  mere  increase 


14  FORMULAE  AND  TABLES  FOR  THE 

in  thickness  beyond  a  certain  point  will  not  appreciably  decrease 
the  resistance,  since  the  current  flow  will  always  be  practically 

limited  to  two  surface  layers  of  thickness  s  =  —• 

m 

Example.  —  Copper  conductor,  p  =  1700. 
For  n  =  100,  s  =  0.67  cm. 

n  =  1000,  s  =  0.21  cm. 

n  =  50,000,  s  =  0.03  cm. 

For  frequencies,  therefore,  of  100,  1000  and  50,000  it  would  be  a 
waste  of  material  to  increase  the  thickness  of  conductors  beyond 
1.34  cm.,  0.42  cm.  and  0.06  cm.,  respectively.  An  additional  in- 
crease in  thickness  of  conductor  would  diminish  the  direct  current 
resistance,  but  it  would  increase  in  the  same  proportion  the  ratio 

7-> 

-5  so  that  the  high-frequency  resistance  would  remain  sensibly 


constant. 

In  the  case  of  iron  conductors,  even  for  comparatively  low  fre- 
quencies the  current  flow  is  limited  to  very  thin  surface  layers  of 
the  conductors. 

Example.  —  Iron  plate,  p  =  104,  /*  =  103. 

For             n  =  50,  s  =  0.07  cm. 

n  =  100,  s  =  0.05  cm. 

n  =  500,  s  =  0.022  cm. 

n  =  1000,  s  =  0.016  cm. 

Resistance  of  Coils  with  Alternating  Currents.  —  The  effect  of 
frequency  on  the  increase  of  resistance  of  coils  is  more  marked 
than  that  of  straight  conductors.  The  problem  has  been  treated 
mathematically  by  several  investigators  and  formulae  have  been 
derived  by  which  the  alternating  current  resistance  of  coils  can  be 
determined  quite  accurately. 

In  formulae  (26)  to  (35)  given  below,  the  following  notation  is 
used. 

a  =  radius  of  coil  in  cm. 

r  =  radius  of  wire  in  cm. 

s  =  number  of  turns  per  cm. 

p  —  specific  resistance  in  C.G.S.  units. 

o>  =  2  TT  X  frequency  =  2  wn. 
RQ  =  continuous  current  resistance. 

R  =  alternating  current  resistance. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     15 

A  very  simple  formula,  but  of  very  limited  application,  has  been 
given  by  Prof.  Wien:* 


0.272 

V     p      / 

This  formula  is  applicable  only  when  the  frequency  is  low  or 
the  radius  of  the  conductor  is  small,  that  is, 


Besides  this  the  coil  must  be  long  compared  with  its  diameter. 
The  length  of  the  coil  should  be  greater  than  seven  times  its 
diameter. 

A  general  formula  derived  by  the  author,  f  which  is  applicable 
for  all  frequencies  and  any  size  of  wire,  provided  only  the  wind- 
ings are  close  together  and  the  coil  is  fairly  long  compared  with  its 
diameter,  is  the  following: 


p  =  P    i  _  x 

P         xTi*2(a*2  +  &2)    cosh4o»r-cos4&r    ; 
where 


16  7T5 


m  =  ~      and      x  =  1,  3,  5,  7,  .  .  .  . 

When  r  is  small  (1  or  2  mm.)  and  the  frequency  is  fairly  high  then, 
to  a  very  high  degree  of  approximation, 


256  s2o>27rra 


(28) 


When  the  frequency  is  high  and  the  radius  of  the  wire  is  not  very 
small,  so  that  we  may  neglect  m4  compared  with  —  ^-,  formula 

*  M.  Wien,  Ann.  d.  Physik,  14,  1,  1904. 

t  Louis  Cohen,  Bulletin  Bureau  of  Standards,  4,  162,  1907. 


16  FORMULAE  AND  TABLES  FOR  THE 

(27)  reduces  to 

-  (29) 

P  ) 

When  the  frequency  is  so  low  that  --  ^  —  may  be  neglected  com- 

pared with  w4,  formula  (27)  reduces  to 

,  107rW>  ,qm 

1  +  --  2  --  >  •  (30) 

Prof.   Sommerfeld  *  derived  the   following  two  formulae  corre- 
sponding to  (29)  and  (30). 

For  very  high  frequencies,  when  ~  V  -    -  >  6, 


(2-«).  (3D 

The  value  of  0  (2  TITS)  has  not  been  determined,  but  from  the 
experimental  results  of  Black  f  the  following  values  of  <£  were 
obtained  : 

2  rs  =  0,  0.1,  0.2,    0.3,    0.4,    0.5,    0.6,    0.7,    0.8,    0.9    1.00 
0  =  1,  1,      1.1,    1.27,  1.45,  1.67,  1.92,  2.2,    2.6,    3.1,  3.7. 

For  low  frequencies, 


(32) 


R  = 


Example.  — 

r  =  0.05  cm.,     w  =  27rX5X105,     s  =  8,     p  =  1700. 
By  formula  (29)  we  get 

/o  _  2  \/  K  \/  i  r>5 

•  =  6.43. 


By  formula  (31)  we  have  for  2  rs  =  0.8,  <£  =  2.6 

p          rj  ric 

/t          U.Uc)  .   .  ^ 


1700 


=7.02. 


*  Sommerfeld,  Ann.  d.  PAi/s.,  24,  609,  1907.     See  also  A.  Essau,  Jahrbuch 
d.  Drahtlosen  Teleg.,  Vol.  4,  490,  1911. 
t  Black,  Ann.  d.  Physik,  19,  157,  1906. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     17 

If  in  the  above  example  we  put  co  =  2  TT  X  5  X  103,  we  may 
apply  formulae  (30)  and  (32). 
By  (30) 

R  107  (125)2  X  IP"12  X  64  X  4  7T2  X  25  X  106  _  1  m7 

RQ  ~  (1700)2 


By  (32) 


?L  -     _u  J7T2  X  47r2  X  25  X  106  X  625  X  IP"8 
£o  "  (1700)  2 

\  +  (2.51)2  +        (2.51)4  +  .  .  .     =  1.036. 


It  is  seen  that  in  the  above  examples  the  agreement  between 
the  two  sets  of  formulae  is  fairly  good.  It  may  be  of  interest  to 
compare  the  difference  in  the  increase  in  resistance  for  a  coil  and 
straight  wire.  Using  the  same  data  as  in  the  above  example, 
r  =  0.05  cm.  and  n  =  5  X  105,  we  have  by  formula  (6)  for  the 
case  of  straight  wire, 

jr  =  r  (ma),  (6') 

m  =  2ir 


0 
r  (ma)  =  0.35355  ma  +  0.25  +  =  2.91, 


hence  =  2.91. 

KQ 

For  the  same  size  of  wire  and  the  same  frequency  the  wire  wound 

73 

into  a  coil,  jr-  —  6.43:  hence  winding  the  wire  into  a  coil  has  more 
tio 

than  doubled  its  high-frequency  resistance. 

When  the  frequency  is  neither  very  high  nor  very  low,  and  the 
windings  are  close  together,  the  general  formula  (27)  should  be 
used. 

Dr.    Nicholson  *   has   derived   an   expression   for   alternating 

*  J.  W.  Nicholson,  Phil.  Mag.,  19,  77,  1910. 


18  FORMULAE  AND  TABLES  FOR  THE 

current  resistance  of  coils  which  is  only  applicable  when  the 
windings  are  at  some  distance  from  each  other,  as  follows: 


(33) 


?L  =     j_   *  r47T2co2  _  J_  /r47T2co2\2 
flo         +  12     P2         180  \    p2    I  ' 


where  a  is  the  angular  pitch  of  the  winding. 

Formulae  (26)  to  (33)  are  applicable  only  for  single  layer  coils. 
For  coils  of  more  than  one  layer  we  have  the  following  formula 
given  by  Wien: 


R  _  2  ( 

-  =  L+-  "  — 


where  TI  and  r2  are  the  inner  and  outer  radii,  I  the  length  of 
the  coil  and  ra  is  the  number  of  layers.  This  formula  holds 
only  for  coils  of  length  at  least  eight  times  the  diameter  and 

when  the  frequency  is  not  very  high,  f3ry-    -<3y- 

For  flat  coils,  length  of  coil  small  compared  with  diameter,  we 
have* 


R       1    .         rT*         S  i    i       3  ri2      )    „  .     . 

=  '1  +  - 


Alternating  Current  Resistance  of  Large  Slot-wound  Con- 
ductors. —  In  the  case  of  alternators,  induction  motors,  etc., 
the  conductors  are  put  into  iron  slots,  that  is,  the  conductor  is 
surrounded  by  iron  on  three  sides.  The  eddy  current  losses  in 
such  conductors  carrying  alternating  currents  are  considerable, 
in  some  cases  many  times  the  PR  loss.  That  is,  the  effec- 
tive resistance  for  alternating  currents,  because  of  the  eddy 
currents,  may  be  several  times  the  resistance  as  calculated  from 
the  dimensions  and  specific  resistance  of  the  conductor. 

This  problem  was  discussed  in  an  interesting  paper  by  Mr. 
Field,  f  who  obtained  a  formula  giving  approximately  the  effective 
resistance  of  the  conductor.  He  also  derived  expressions  for  the 

*  A.  Essau,  Jahrbuch  der  Drahtlosen  Telegraphic  und  Telephonic,  Vol.  4, 
p.  490,  1911.  , 

t  A.  B.  Field,  Proc.  A.I.E.E.,  24,  659,  1905. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     19 

current  distribution,  density  and  phase,  in  the  entire  sections  of 
the  conductors. 

If  we  denote  by  RQ  the  direct  current  resistance  and  by  R  the 
alternating  current  resistance,  we  have  for  the  mth  layer 

RQ 

A  (m2  —  m)  (cosh  af—  cos  af)  (sinh  af—  since/)  -f-  (sinh  2  af  —  sin  2  af) 
cosh  2  af  —  cos  2  af 

(36) 


FIG.  4. 


FIG.  5. 


where 


/  =  depth  of  conductor  in  cm.  (if  a  laminated  conductor  take 
the  gross  depth). 


a 


0.145  V. 


n  =  frequency. 

r2  =  practically  unity  for  solid  conductors,  and  for  laminated 
conductors  is  generally  equal  to  the  ratio  of  half  the 
mean  length  of  turn  to  gross  length  of  core. 

r\  =  for  solid  conductors,  the  ratio  of  net  copper  measured 
across  the  slot,  to  the  slot  width,  and  for  laminated  con- 
ductors generally  the  same  multiplied  by  the  ratio  of  net 
to  gross  conductor  depth. 


20  FORMULAE  AND  TABLES  FOR  THE 

m  denotes  the  mth  layer,  counting  the  number  of  layers  from 
the  bottom  of  the  slot  upwards. 

RQ 

To  facilitate  the  computation  of  -5-  as  given  by  equation  (36) 

K 

Mr.  Field  worked  out  a  set  of  curves,  which  are  reproduced  in 

n 

Fig.  6,  giving  the  values  of  -^-  for  different  values  of  af  and  m. 

As  an  illustration  consider  the  problem  of  a  two-layer  winding 
having  the  following  dimensions: 
Slot  width  =  0.65  in. 
Conductors,  2-1  (0.4  X  0.75  in.). 

7-2  =  1,     n  =  ~  =  0.615,    /  =  0.75  X  2.54  =  1.905. 
U.oo 

For  25  cycles,  a  =  0.145  \/25  X  0.615  =  0.568,  af  =  1.08. 
For  60  cycles,  a  =  0.145  A/60  X  0.615  =  0.88,  af  =  1.68. 
From  the  curves  given  in  Fig.  6  we  obtain  the  following  values 

t      R 
for  jc-: 

25  cycles. 

7-» 

Bottom  conductor,     m  —  1,      ^-  =  1.12. 

/to 

Top  conductor,          m  =  2,     TT  =  2.00. 

£lQ 

60  cycles. 

r> 

Bottom  conductor,     m  =  1,     "5"  =  1-55. 

R 

Top  conductor,          m  =  2,      -5-  =  5.6. 

/to 

Mr.  Field  discusses  at  some  length  the  relative  advantages  of 
different  form  of  windings  and  the  original  paper  should  be  con- 
sulted by  those  who  are  particularly  interested  in  this  subject. 

Distribution  of  Induction  and  Eddy  Currents  in  a  Long  Iron 
Pipe,  Enclosing  a  Conductor  Carrying  an  Alternating  Current.  — 
It  is  now  generally  appreciated  that  single  core  cables  designed  to 
carry  alternating  currents  cannot  be  laid  in  separate  iron  pipes,  on 
account  of  the  large  eddy  current  and  hysteresis  losses  which  will 
be  produced  by  the  magnetic  induction  in  the  iron  pipes.  It  is 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     21 


Values  of  af 

2.7       2.6      2.5       2.4      2.3       2.2      2.1      2.0      1.9       1.8      1.7        1.6      1.6      1.4 


1.0 


0       0.1      0.2       0.3      0.4       0.5       0.6      0.7      0.8       0.9      1.0      1.1       t2       1.3 

Values  of  af 
FIG.  6. 


1.4 


22  FORMULAE  AND  TABLES  FOR  THE 

interesting,  however,  to  know  what  the  energy  loss  is  likely  to  be  in 
any  particular  case,  and  we  give  here  the  formulae  as  developed  by 
Mr.  Field*  for  the  flux  distribution  and  energy  loss  in  iron  pipes 
enclosing  single  core  cables  which  carry  alternating  current. 
Let 

I  =  current  amplitude  in -cable, 

h  =  thickness  of  pipe  wall  in  cm., 

d  =  internal  pipe  diameter  in  cm., 

L  =  mean  circumferential  length  of  pipe  in  cm., 

JLI  =  mean  permeability, 
4717,1 

'— uT.- 

m=  0.0002V/—, 
p 

w  =  2  TT  X  frequency  =  2  7m, 
p  =  specific  resistance  (electrical)  iron, 
p  _  Im  .  /cosh  mh  —  sin  mh 
L  V  cosh  mh  +  cos  mh ' 
Bx  =  induction  in  lines  per  sq.  cm.  at  point  distant  x  from 

surface, 
I9  =  current  density  in  amperes  per  sq.  cm.  of  the  longitudinally 

flowing  current  at  point  distant  x  from  surface, 
W  =  eddy  current  loss  per   cm.  length  of  pipe  expressed  in 

watts. 
Two  cases  are  considered: 

(1)  Where  the  pipe  is  laid  in  dry  concrete  or  dry  earth,  so  that 
no  eddy  currents  leak  out  or  leave  the  pipe. 

(2)  Where  the  pipe  is  laid  in  wet  soil,  or  where  the  ends  are  for 
any  other  reason  at  the  same  potential. 

CASE  1 

Ix  =  Plem*sm(a>t  +  mx  +  $)-em(h-*)s™[ut+m(h-x)  +  <l>]\.     (37) 
The  amplitude  of  Ix  may  be  put  in  the  following  form: 


7   _  -^m  . /2  f  cosh  m  (h  —  2  x)  —  cos  m  (h  —  2  x)  j 
x  ~  L  V  cosh  mh  +  cos  mh 

*  M.  B.  Field,  UA  theoretical  consideration  of  the  currents  induced  in  cable 
sheaths  and  the  losses  occasioned  thereby,"  Journal  Inst.  of  Elect.  Eng.,  Vol.  33, 
pp.  936-963,  Apr.,  1904. 


_  If]  *  / 

x  ~~  L  V 


...  , 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     23 

For  the  flux  distribution  we  have 

Pm  X  108  p  c  .  ,  , 

Bx  = \  emx  (sin  —  cos)  (cot  +  mx  -f-  <f>) 

CO 

-f  em  (*-*)  (sin  -  cos)  [coZ  +  m  (h  -  x)  +  0] } ,          (39) 
and  the  amplitude  of  Bx  is 

cosh  m  (A  —  2  ap  +  cos  m  (h  —  2  a?)  _  ,.„, 

cosh  mh  +  cos  raft 

The  energy  loss  is 

w  _  Pmp  sinh  mh  —  sin  mh 
L  '  cosh  mh  +  cos  raft 

CASE  2 
We  have 

.^  _  Pmp  sinh  2  raft  —  sin  2  raft 
2L  cosh  2  raft  +  cos  2  mh 

Jx  and  #3;  are  of  the  same  form  as  (38)  and  (40)  except  that  we 
write  2  h  in  place  of  h. 

Example.  — 

d=2cm.,    h  =  0.1  cm.,    7  =25 amp.,   p  =  27rX60,    ^=1000, 

TJ  =  ~2  =  1256,     p  =  10~5  ohm,      L  =  TT  (d  +  h)  =  6.6  cm. 
m  =  15.4,     raft  =  1.54. 

sin  2  raft  =  0.061,        cos  2  raft  =  -  0.998, 
sinh  2  raft  =  10.86,      cosh  2  raft  =  10.9, 

w     625  X  15.4 X10~5  vx  10.86-0.061 

-^T      -  X  10.91-0.998  =  159°  X 10    WattS  Per  Cm* 

If  the  pipe  is  200  feet  long,  the  energy  loss  is 

W  =  1590  X  10~5  X  200  X  30.5  =  97  watts. 

For  100  amperes  under  the  same  conditions  the  loss  is 
W  =  16  X  97  =  1552  watts. 

Effect  of  Grounded  Lead  Covering  of  Single  Conductor  Cable. 

—  The  lead  covering  acts  as  a  short-circuited  secondary  to  the 
e.m.f .  generated  by  the  inductive  reactance  of  the  line.  It  reduces 
therefore  the  inductance  of  the  line,  but  increases  the  energy  loss 
due  to  the  currents  in  the  lead  covering. 


24  FORMULAE  AND  TABLES  FOR  THE 

Let 

R  =  line  resistance, 

Ri  =  resistance  of  lead  covering  (for  the  same  cross-section,  four- 
teen times  that  of  copper), 

x0  =  line  inductive  reactance  if  no  lead  covering  were  used 
and  a  conductor  of  the  same  outside  diameter  as  that  of 
the  lead  covering  were  used  in  the  calculations. 

The  formula  obtained  by  Berg*  for  the  effective  resistance  and 
effective  reactance  are  as  follows: 


Effective  reactance  = 


As  a  rule  XQ  is  small  compared  with  Ri  and  we  can  write, 

(43) 


/Y»      2 

Effective  resistance  =  R  +  -5- 


Effective  reactance  =  z0. 

To  calculate,  therefore,  the  effective  resistance  and  effective 
reactance  of  a  lead  covered  cable,  we  determine  the  resistances  of 
conductor  and  resistance  lead  covering  R  and  Ri,  and  x0  =  2irnL. 
The  inductance  L  is  taken  as  that  between  conductors  of  the  same 
diameter  as  the  outside  diameter  of  the  lead  covering  and  the  dis- 
tance, the  interaxial  distance  between  the  conductor  and  its  return. 

Example.  —  Two  No.  0000  B.  &  S.  conductors  each  enclosed  in 
a  lead  sheath  of  external  diameter  =1"  and  internal  diameter 
=  0.75".  Interaxial  distance  between  the  conductors,  D  =  4.5". 
Length  of  conductors  =  1000  feet.  To  determine  the  effective 
resistance  and  reactance  at  25  and  60  cycles. 

We  have  R  =  0.049  ohm, 

area  of  lead  =  J  TT  X  f  =  0.343  sq.  in., 
and  Ri  =  0.332  ohm. 

At  25  cycles,  we  have  for  the  inductive  reactance, 
xQ  =  2  TT  X  25  (2  loge  j  +  O.s)  ^^y  *'"  X  ^g  =  0.0234  ohm. 

*  Ernst  J.  Berg,  Constants  of  Cables  and  Magnetic  Conductors,  Proc. 
A.I.E.E.,  April,  1907. 


CALCULA T10N  OF  ALTERNATING  CURRENT  PROBLEMS     25 


Therefore, 


Rett  =  0.049  + 


(0.0234)' 
0.332 


=  0.0506  ohm. 


For  a  frequency  of  60  cycles  we  have 

XQ  =  0.0561  ohm 
and  R^  =  0.0574  ohm. 

The  resistance  is  increased  by  3  per  cent  for  the  25  cycle  fre- 
quency and  by  16  per  cent  for  the  60  cycle  frequency. 

The  table  given  below  applies  to  a  single-conductor  cable  No. 
0000  B.  &  S.  having  a  lead  covering  TVin.  thick  with  an  outside 
diameter  of  1  in. 

FREQUENCY  =  25. 


Interaxial  distance  in  inches. 

2.13 

4 

12 

24 

R 

0  264 

0.264 

0.264 

0.264 

2*0 

0.0497 

0.916 

0.1544 

0.1904 

a-o2                             •    • 

0.0025 

0.0084 

0.0238 

0.0362 

Rett                       

0.2651 

0.2676 

0.2743 

0.2796 

Increase     .       

0.42% 

1.37% 

3.9% 

5.9% 

FREQUENCY  =  60. 


Interaxial  distance  in  inches. 

2.13 

4 

12 

24 

R                           

0.264 

0.264 

0.264 

0.264 

XQ                                     

0.1193 

0.220 

0.3706 

0.4570 

Z02                

0.0142 

0.048 

0.137 

0.208 

7?efl    

0.2701 

0.2848 

0.3233 

0.3536 

Increase  

2.3% 

7.9% 

22.5% 

33.9% 

Effect  of  Ungrounded  Lead  Covering  in  a  Single-conductor 
Cable.  —  Professor  Berg  *  has  also  derived  expressions  for  the  eddy 
current  and  energy  losses  in  an  ungrounded  lead  covered  cable, 
as  follows: 


Eddy  currents  in  lead     =  j-^- 
Energy  loss  in  lead 


4ft 


(44) 


Effective  resistance,  Reft  =  R  + 
*  1.  c.,  p.  24. 


co2L2 


26 


FORMULAE  AND  TABLES  FOR  THE 


where       co  =  2  IT  X  frequency, 

.Ri  =  lead  resistance  in  ohms, 
R  =  resistance  of  conductor  proper  in  ohms, 

inductance  per  mile  in  henrys, 


322 

L  =  --j  loge 


T 

D  =  interaxial  distance  between  the  conductor  and  its 

return, 
DI  =  external  diameter  of  lead  sheath. 


As  an  illustration  we  may  use  the  same  data  as  given  in  the 
example  for  the  case  of  grounded  lead  covered  cable. 
We  have  then 

D  =  4.5  in.,    Z>i  =  1  in., 


475 


hence, 


R  =  0.049  ohm, 
R!  =  0.332  ohm, 
w  =  27rX60, 


=  0.049 


=  a()49 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     27 

It  is  seen  from  this  example  that  the  increase  in  resistance  is 
negligible  if  the  lead  covering  is  not  grounded. 

Leakage  Conductance.* — All  insulating  materials  are  more  or 
less  conducting  to  some  extent;  there  is  no  absolutely  perfect 
insulator.  Hence,  when  a  difference  of  potential  exists  between 
two  conductors  which  are  embedded  in  an  insulating  medium,  as 
in  the  case  of  the  cores  in  a  cable,  there  is  a  current  flowing 
from  one  conductor  to  the  other  through  the  insulating  material 
separating  the  two  conductors.  In  consequence  of  the  incomplete 
insulation  and  also  because  of  dielectric  hysteresis  and  electro- 
static radiation,  we  have  an  energy  component  of  the  current, 

I=Eg, 

which  is  in  phase  with  the  e.m.f. 

We  give  here  some  formulae  for  the  calculation  of  the  leakage 
conductance  of  linear  conductors. 

The  leakage  conductance  between  a  cylindrical  conductor  and 
an  infinite  conducting  plane  with  uniform  distance  d  between  the 
plane  and  the  axis  of  the  cylinder  is 

(45) 


where 

a  =  radius  of  cylinder  in  cm., 

d  =  distance  between  plane  and  axis  of  cylinder  in  cm., 
I  =  length  of  cylinder  in  cm., 
p  =  specific  resistance  of  the  medium  in  C.G.S.  units. 

The  resistance  between  a  cylinder  and  an  infinite  plane  is  the 
reciprocal  of  the  conductance  given  by  equation  (45). 

,      2d 
ploge  — 

' 


*  See  "  Theorie  der  Wechselstrome,"  von  J.  L.  La  Cour  und  O.  S.  Bragstad, 
Zweite  Auflage,  pp.  588-596. 


28  FORMULAE  AND  TABLES  FOR  THE 

Example.  — 

a  =  3  cm.,  d  =  20  cm., 

P  =  5  X  1010,        I  =  100  cm., 

2d  40 

log,--  =  log,  -o-*  2.59, 

a  o 

„      5X10l°X2.59  , 

^  =  —  o  —  x/  inn  —  =  2  X  1010  absohms, 
ATT  /\  1UU 

10~10 

g  =     0    absmhos. 
z       . 

cm 

If  we  express  p  in  megohms  (106  ohms)  per  -  —  -2  and  I  in  miles, 

cm. 

formula  (45)  reduces  to 


Prof.  Kennelly*  derived  formulae  for  the  leakage  conductance  of 
linear  conductors  in  terms  of  antihyperbolic  functions.  For  the 
formula  corresponding  to  (45),  he  gives 

2-irl  1.01  /yl-x 

g  =  -  -  =  -  -t  mho  per  mile.  (47) 

p  cosh'1  -      p  cosh"1  -  % 

Example.  — 


OTY1 

p  =  5  X  108  megohms  per  —  -2,  approximately  the  value  for  gutta- 

cm. 

percha, 

log,  —  =  log«  10  =  2.302,     cosh-1  ^  =  2.292. 

By  formula  (46)  we  have 


By  formula  (47),  we  get  the  approximately  same  result, 
•-  530^29  =  88X10-  mho  per  mile. 
*  Proceedings  Am.  Phil.  Socy.,  Vol.  48,  pp.  142-165,  1909. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     29 

For  two  parallel  conductors  separated  by  a  distance  d,  and  of 
equal  radii  a,  we  have 

1.01 1 


d+ 


2a       I 


1.01 1  ,,m 
-7  mho,  approx.,                              (49) 


2  P  loge  - 

1.01 1  ,KA. 

or  g  = ^—  (50) 

2  p  cosh"1  TT— 
,4  d 

When  -  is  small,  formula  (49)  is  considerably  in  error. 
As  an  example  suppose  we  take  -  =  2.5.     Then 
log.-  =  0.916, 


log,-  -      -=0.693, 

cosh-1-  =  0.689. 
a 

From  these  values  it  is  obvious  that  (48)  and  (50)  agree  closely 
while  (49)  is  in  error  about  35  per  cent. 

For  concentric  cylinders,  we  have  for  the  leakage  conductance 

1.01  Z 
g=  -  -^mho,  (51) 


where  I  is  the  length  of  cylinders  expressed  in  miles,  p  is  specific 
resistance  expressed  in  megohms,  D  is  the  inner  diameter  of  outer 
cylinder,  and  d  is  the  outer  diameter  of  inner  cylinder. 

In  the  above  conductance  formulae  it  was  assumed  that  the  in- 
sulation is  homogeneous,  having  the  same  resistivity  for  the  entire 
depth  of  the  insulation.  If  this  is  not  the  case  the  computations 
are  very  complex.  We  may  obtain  approximate  results  by  as- 
suming that  the  insulation  consists  of  several  layers  of  different 
resistivities,  and  in  that  case  the  formula  for  concentric  cylinders 
is  as  follows: 


30 


FORMULAE  AND  TABLES  FOR  THE 
1.01 1 


mho, 


(52) 


where  dx  is  the  diameter  of  the  xih  insulation  layer. 
For  a  three-core  cable  we  have 


Q  = 


2.02  I 


mho, 


(53) 


FIG.  8. 


where 

D  =  diameter  of  sheath, 
a  =  radius  of  each  core, 
d  =  interaxial  distance  between  cores. 


Magnetic  Flux  Distribution  and  Eddy 
Current  Losses  in  Iron.  —  It  is  well 
known  that  in  subjecting  iron  to  an 
alternating  magnetizing  force,  eddy  cur- 
rents are  generated  in  the  iron  in  planes 
perpendicular  to  the  direction  of  the 
magnetic  flux.  These  eddy  currents 

produce  magnetic  fields  of  their  own  which  prevent  the  mag- 
netic flux  from  penetrating  completely  into  the  body  of  the 
iron.  As  a  consequence  we  have  an  uneven  distribution  of  the 
magnetic  flux  in  the  iron  and  also  energy  losses.  Both  of  these 
effects  depend  on  the  permeability  and  conductivity  of  the  iron 
and  on  the  frequency.  The  problem  of  determining  the  magnetic 
flux  distribution  and  eddy  current  losses  offers  considerable  math- 
ematical difficulties,  and  it  has  only  been  worked  out  for  a  few 
special  cases  which  we  shall  consider  here. 

Infinite  Plates.*  —  Consider  an  iron  plate,  Fig.  9,  whose  sur- 
face in  the  direction  perpendicular  to  the  plane  of  the  paper  is 
infinite  in  extent,  its  thickness  being  2  a.  Assume  m.m.f.  acting  on 
the  plate  producing  a  magnetic  field  in  the  direction  perpendicular 
to  the  plane  of  the  paper,  and  let  Ba  cos  at  denote  the  magnetic 
flux  density  on  the  surface  of  the  plate,  which  is  produced  by  the 
impressed  m.m.f.  The  magnetic  flux  density  at  any  point  in  the 

*  See  "  Theory  and  Calculation  of  Transient  Electric  Phenomena  and  Oscil- 
lations "  (Chapter  VI)  by  C.  P.  Steinmetz,  and  "  The  Theory  of  Alternat- 
ing Currents,"  Vol.  I,  Chap.  XVI,  by  Alexander  Russell. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     31 


direction  perpendicular  to  the  surfaces,  and  at  distance  x  from 
the  center  is  given  by 


Bx  =  Bf, 


cosh  2  mx  -\-  cos  2  mx 


cos  (at  - 


cosh  2  ma  +  cos  2  ma 

sinh  m(a  —  x)  sin  m  (a+ff)  -j-sinhm  (q-j-a)  sinm  (a  —  x) 
cosh  m  (a— z)cosm(a+a;)-|-coshm  (a+z)cosm  (a— x) 


tan  7s  = 


where  m  =  2 

At  =  permeability, 
n  =  frequency, 

p  =  specific    resistance    in   C.G.S. 

units.  jf. 


(54) 
,   (55) 


From  equations  (54)  and  (55)  it  is 
obvious  that  the  amplitude  of  B 
diminishes  from  the  surfaces  to  the 
center  of  the  plate,  and  the  phase  of 
the  magnetic  flux  changes  from  point 
to  point  along  the  axis  perpendicular 
to  the  surfaces. 

The  minimum  value  of  B  is  at  the  center  of  the  plate; 

BnV2 


FIG.  9. 


tan  7C 


Vcosh  2  ma  +  cos  2  ma 

sinh  ma  sin  ma 
cosh  ma  cos  ma 


(56) 
(57) 


The  mean  or  average  value  of  the  magnetic  flux  density  for  the 
entire  section  of  the  plate  is 


„ 

m 


B 


cosh  2  ma  —  cos  2  ma   a 


V2  am  I  cosh  2  ma  +  cos  2  ma ) 
When  2  ma  is  large,  equation  (58)  reduces  to 


(58) 


(59) 


32  FORMULAE  AND  TABLES  FOR  THE 

The  total  magnetic  flux  in  the  plate  is 

Ba  V2  (  cosh  2  ma  —  cos  2  ma  )  2 


Bt 


m      )  cosh  2  ma  +  cos  2  ma 


(60) 


If  the  flux  density  had  been  uniform  throughout  the  entire 
section  of  the  plate  and  of  an  intensity  equal  to  Ba,  then  the 
thickness  of  plate  2  d  required  to  get  the  same  total  magnetic 
flux  as  that  given  by  equation  (60)  would  have  been 


2d  = 


2      (  cosh  2  ma  —  cos  2  ma  ^ 
m  V2  t  cosh  2  ma  +  cos  2  ma ) 


(61) 


The  value  of  d  as  given  by  equation  (61)  may  be  considered  as 
the  equivalent  depth  of  uniform  magnetic  flux  density. 
When  2  ma  is  large,  equation  (61)  reduces  to 


d  = 


1 


m  V2      27r  V  2fj,n 


(62) 


Example. 


a  =  0.025  cm.,     /*  =  2000,     p  =  104,     n  =  60. 

4  /2  X  103  X  60 
m  =  27rV TnH~ 


=  21.74. 


X, 

cosh  2  mx. 

cos  2  mx. 

B* 

Btl' 

0.000 

1.000 

1.000 

0.972 

0.005 

1.023 

0.976 

0.973 

0.020 

1.403 

0.644 

0.985 

0.025 

1.651 

0.465 

1.000 

The  equivalent  depth,  d  =  0.0244  cm. 

For  such  fine  lamination  and  low  frequency,  the  penetration  of 
magnetic  flux  density  is  practically  complete.  Suppose,  however, 
that  in  the  above  example  we  assume  a  frequency  of  500;  then, 

m  =  62.83. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     33 


X. 

cosh  2  mx. 

cos  2  raz. 

Bx 
Ba 

0.000 

1.000 

1.000 

0.43 

0.005 

1.205 

0.809 

0.44 

0.010 

1.896 

0.309 

0.46 

0.015 

3.368 

-0.309 

0.54 

0.020 

6.230 

-0.809 

0.72 

0.025 

11.590 

-1.000 

1.00 

The  equivalent  depth  d  is  0.012  cm.  which  is  about  one-half  the 
actual  thickness  of  plate. 

The  energy  loss  due  to  the  eddy  currents  is  given  by  the  ex- 
pression 

sinh  2  ma  —  sin  2  ma  )  ?    watts 


W 


X10~7 


cosh  2  ma  +  cos  2  ma  \    cu.  cm. 


(63) 


When  ma  is  greater  than  TT,  that  is  2  y  —  a  greater  than  unity, 

the  factor  in  the  brackets  of  equation  (63)  is  practically  unity, 
and 

rap#o2  X  10-7 


W  = 


cu.  cm. 


HQ  being  the  magnetizing  force. 

When  ma  is  small  the  expression  for  W  as  given  by  (63)  is 
approximately  as  follows: 

K2  ma)3 


X 


wV#o2  X  10-7    watts 


24  7T2 


cu.  cm. 


approximately.          (65) 


For  small  values  of  ma  the  magnetic  flux  is  practically  con- 
stant through  the  entire  section  of  the  iron  plate,  and  hence  we 
may  assume  constant  permeability.  Hence  putting  #max  =  nH0) 
equation  (65)  becomes 

watts 


.  1.64(2^ 


10-' 


cu.  cm. 


In  the  above  equation  the  thickness  of  the  plate  2  a  must  be  ex- 
pressed in  cm.  p  is  the  resistivity  in  C.G.S.  units  (about  10,000), 
n  is  the  frequency  and  B  is  the  maximum  value  of  the  flux  density. 


34 


FORMULAE  AND  TABLES  FOR  THE 


The  values  of  equivalent  depths  d  (of  uniform  magnetization) 
in  cm.  for  various  materials  and  frequencies  are  given  below:* 


Frequency. 

25 

60 

1000 

10,000 

10« 

Soft  iron,  M=  103,  P  =  104  
Cast  iron,  /*  =  200,  p  =  105  

0.0714 
0.504 

0.922 

7.14 

0.0460 
0.325 

0.595 
4.60 

0.0113 
0.080 

0.144 
1.13 

0.0036 
0.0252 

0.0461 
0.357 

0.00036 
0.0025 

0.0046 
0.036 

i  rv4 
Copper,      u  =  l,       p  =  — 

Resistance  alloys,  n  =  1,    p  =  105  .  .  . 

Iron  Cylinders.  —  The  problem  of  magnetic  flux  distribution 
and  eddy  current  losses  in  iron  cylinders  has  been  discussed  by 
several  writers,  and  the  results  have  been  summarized  in  a  very 

APPROXIMATE  FORMULA 


Function  . 

x<  1. 

x   >  6. 

ber  (x) 
bei(x) 

X(x} 
Y(x) 

Zf     \ 
(X) 

W(x) 

0  (z) 
d  (z) 

w(x) 

r4 

1          *             I          n  01  ^9^  -r-4 

r>r»a  /?  •   /v        O  7O71  -r 

0.046 

64 

T2/              T4   \         T2 

11            I         (i      n  nni7^ft  T**^ 

V27TZ 

0.0884      0.046 

fa 
cin  ft'   R       (\  7O71  r 

4  \       576/      4 
1  +  ^  =  1  +  0.03125  z4 

T2/               T4   \         T2 

ill     '       I             ^1  4-  0  OO'i0!  T4"\ 

V2wx 
0.0884      0.0625 

x             x2 
^(,       0.7071  ,  0.25  ,  0 

z3 
.265\ 

4  V        192/      4 

.V1            x           z2 

z3    j 

x(  1  -i-  ^  \     ^  n  -u  n  m  04  r4<i 

Y  ^07071  1  a°88  1  al25 

2V1  +  96j      2(1 
!_  J|Z4  =  1-0.01302  z4 

~2/              T4  \         3-2 
fl                  1             /I        O  OO7S1  i-4N 

2      0.7071      0.25 

z          z2           z3 
0.7071      0.25 

si1      128J      8  (1 
zV         11      \    z2 

2.8284      2      0.3536 

z4 

z          z2         z3 
1.4142      0.1768 

48-1 

z              z3 

*  This  table  is  taken  from  Steinmetz  "  Trans.  Electrical  Phenom.,"  p.  365. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     35 

interesting  paper  by  Prof.  P.  O.  Pedersen,  which  appeared  in  the 
Jahrbuch  der  Drahtlosen  Telegraphic  und  Telephonic,  Vol.  4,  pp. 
501-515.  He  also  worked  out  very  valuable  tables  to  facilitate 
numerical  computations.  The  formulae  and  tables  given  in  this 
section  are  taken  from  that  paper. 

In  the  formulae  given  in  this  section,  certain  functions  are 
employed  which  are  extensions  of  the  ber  and  bei  functions  first 
introduced  by  Dr.  O.  Heaviside  and  Lord  Kelvin.  The  values  of 
all  these  functions  for  arguments  from  0  to  6  are  given  in  Table  II, 
p.  43.  For  arguments  greater  than  6  or  less  than  unity  the  values 
of  the  functions  are  given  by  the  foregoing  approximate  expres- 
sions, the  largest  error  in  these  approximate  formulae  being  less 
than  1  per  cent. 

Magnetic  Flux  Distribution  in  Iron  Cylinders.  —  Assume  an 
iron  cylinder  subjected  to  an  uniform  alternating  m.m.f.  in  the 
direction  parallel  to  the  axis  of  the  cylinder.  This  can  be  realized 
in  practice  by  placing  the  iron  cylinder  in  a  long  coil  through 
which  alternating  current  is  flowing. 

Let    a  =  radius  of  cylinder  in  cm., 

A  =  sectional  area  of  cylinder  in  sq.  cm., 
H  =  permeability, 

p  =  specific  resistance  in  C.G.S.  units, 
n  =  frequency, 
co  =  2irn, 


P 
/o  cos  a)t  =  current  in  coil, 

N  =  number  of  turns  per  cm., 


ir 
Ha  =          IQ  (magnetizing  force),  the  intensity  of  magnetic 

field  at  the  surface  of  cylinder. 

The  intensity  of  the  magnetic  field  at  any  point  distant  r  from 
the  axis  of  the  cylinder  is  given  by 

tfr  =  tfrcosM-7r),  (67) 


_  ber  (mr)bei  (ma)  —  bei  (mr)  ber  (ma) 
a  7r  ~  ber  (mr)  ber  (ma)  +  bei  (mr)  bei  (ma) 


11 


:::  3 


n 


I,  El 


II 


and  tan  5  =  5  (ma).  ] 

The  density  of  the  eddy  currents  at  any  point  distant  r  |; 
axis  of  the  cylinder  is  given  by 

Jr  =  IT  COS  (ut  -  ]8r), 

10 


where 


and 


jY(mr) 


tanfr 


X(ma) 
ber'  (mr)  bei  (ma)  —  bei'  (mr) 


ber'  (mr)  ber  (ma)  +  bei'  (mr)  bei  (ma) 
At  the  axis  of  the  cylinder 


/o  =  0    and    tan  /30  =  — 


ber  (ma) 


bei  (ma) 

The  energy  loss  due  to  the  eddy  currents  in  one  centime 
of  cylinder  is 


fJLH 


AHa2w  (ma)  watts. 


8X109 

The  increase  in  resistance  of  the  coil  due  to  the  dissi 
energy  in  the  iron  cylinder  or  core  in  one  centimeter 
cylinder  is 

Ru  =  — r^r  AN2w  (ma)  ohms. 

The  inductance  per  centimeter  length  of  coil  is 

4?r 
Lu  =  —  nAN2S  (ma)  henry. 


(76) 


The  increase  in  inductance  of  coil  per  centimeter  length  due  to 
presence  of  iron  cylinder  within  the  coil  is  given  by 


AL 


(77) 


Equations  (67)  and  (77)  give  all  the  necessary  formula  for  a  com- 
plete study  of  magnetic  flux  distribution,  energy  losses,  etc.,  in 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     37 


an  iron  cylinder  placed  in  a  variable  magnetic  field  whose  direc- 
tion is  parallel  to  the  axis  of  the  cylinder. 

We  shall  now  consider  a  few  examples  in  illustration  of  the 
above  formulae. 

Example.  —  An  iron  cylinder  is  subjected  to  an  alternating  mag- 
netizing force  of  intensity  Ha  in  the  direction  parallel  to  its  axis. 
What  is  the  intensity  of  magnetizing  force  and  density  of  eddy 
currents  at  any  point  along  its  radius? 

Let  a  =  0.5  cm., 

v  =  103  (assumed  constant  for  entire  section  of  cylinder), 
p  =  104  C.G.S.  units, 
n  =  25  cycles  per  second. 

We  have  m  =  14,      

ma  =  7,     VX  (ma)  =  22.4. 


r. 

Vx  (mr). 

Hr 

Ha' 

I, 
Ha' 

Vy(mr). 

0.50 

22.4 

21.28 

1 

10.58 

0.45 

13.86 

13.10 

0.62 

6.51 

0.40 

8.98 

8.44 

0.40 

4.19 

0.35 

5.90 

5.50 

0.26 

2.73 

0.30 

3.87 

3.57 

0.17 

1.77 

0.20 

1.75 

1.61 

0.08 

0.70 

0.10 

1.06 

0.71 

0.047 

0.35 

0.00 

1.00 

0.00 

0.044 

0.00 

From  an  inspection  of  the  fourth  column  in  the  above  table  it  is 
seen  that  for  an  iron  cylinder  of  0.5  cm.  diameter  even  for  such  low 
frequency  as  25  cycles,  the  magnetization  diminishes  very  rapidly 
as  we  recede  from  the  surface  of  the  cylinder.  This  shows  that 
even  for  low  frequencies  lamination  is  required  to  obtain  complete 
penetration  and  not  make  the  eddy  currents  excessive. 

By  formula  (70)  the  total  magnetic  flux  in  the  cylinder  is 

$o  =  TT  (0.25)2  X  108£Ta0  (ma), 
<j>  (ma)  =  0.272, 
$o  =  53.3  Ha. 

For  zero  frequency  <f>  (ma)  =  1 ;  hence  the  effect  of  the  frequency 
which  prevents  complete  penetration  is  to  reduce  the  flux  to 
about  27  per  cent  of  its  value  for  steady  magnetization. 

Referring  to  formula  (70),  it  is  evident  that  if  there  were  no 


38  FORMULAE  AND  TABLES  FOR  THE 

iron  present,  the  total  flux  would  be  $  =  AHa;  hence  we  may 
consider  the  factor  /*<£  (ma)  as  the  equivalent  permeability  of 
the  iron  for  the  variable  magnetizing  force.  Now  it  may  happen 
that  when  the  frequency  is  high  and  the  radius  of  the  cylinder 
is  not  very  small  the  quantity  /x</>  (ma)  would  be  less  than 
unity,  which  would  give  the  effect  of  permeability  less  than  unity. 
This  would  mean  that  the  penetration  is  limited  to  such  thin  sur- 
face layers  of  the  cylinder  that  the  total  flux  is  less  than  what  it 
would  be  if  the  iron  were  absent  and  the  flux  were  passing  through 
the  air  space  occupied  by  the  cylinder.  For  instance  in  an  iron 
cylinder  of  1  cm.  radius  subjected  to  alternating  magnetizing  force, 
assuming  ju  =  103,  p  =  104,  we  have  for 

n=      0       100      103      104      5.104      105      106 
ji0  (ma)    =  1000       70       22         7         3.2      2.2      0.7 

In  this  particular  example  at  the  frequency  of  one  million  cycles  per 
second  the  total  flux  is  less  than  if  the  iron  were  entirely  removed. 

Let  us  now  consider  a  more  practical  case  —  that  of  a  sole- 
noidal  coil  having  a  core  made  up  of  a  bundle  of  iron  wires.  We 
shall  find  the  total  flux  in  the  core,  the  energy  loss  and  the  in- 
crease in  resistance  of  coil  due  to  energy  dissipation  in  the  iron 
core. 

We  shall  assume  a  coil  10  cm.  in  diameter  and  15  cm.  long, 
total  number  of  turns  60,  or  4  turns  per  centimeter,  the  frequency 
60  cycles  per  second  and  the  current  in  the  coil  2  amperes.  The 
core  is  made  up  of  1000  iron  wires,  insulated  from  each  other,  the 
radius  of  each  wire  being  0.1  cm.  We  shall  also  assume  a  constant 
permeability  of  1000  for  the  entire  section  of  each  wire. 

The  magnetizing  force 


/2  X  60  X  103 


•I-2*!/-  --  =  21.74. 


The  radius  of  each  wire  being  0.1  cm., 

ma  =  2.174. 

From  Table  II,  page  43,  we  find 

<t>  (ma)  =  0.8062, 
w  (ma)  =  0.728. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     39 

The  flux  for  one  wire  of  the  core  is 

$o  =  TT  (O.I)2  X  103  X  10.05  =  254.5 
The  total  flux  through  the  entire  core  is 

$  =  254.5  X  1000  =  254,500. 
The  energy  loss  due  to  eddy  currents  in  one  wire  is,  by  formula  (74), 

ins  y  fin 

"       °    TT  (10.05)2  X  0.728  X  15  =  0.026  watt. 
o  X  lir 

The  loss  of  energy  in  the  entire  core  is 

W  =  0.026  X  103  =  26  watts. 

The  increase  in  resistance  of  coil  due  to  eddy  current  losses  in  one 
wire  is,  by  formula  (75), 

ft.  =  2TX1°327rX60  <r(0.1)2X  16X0.728X15  =  0.013  ohm. 


The  increase  in  resistance  due  to  eddy  current  losses  in  the 
entire  core  is 

R»  =  0.013  X  103  =  13  ohms. 

Toroidal  Coils.  —  Magnetic  flux  distribution  in  an  iron  core  of 
a  toroidal  coil. 

In  the  derivation  of  the  formulae  given  below  it  is  assumed  that 
the  core  is  made  up  of  a  number  of  very  thin  plates,  insulated 
from  each  other,  the  thickness  of  the  plates  being  very  small  com- 
pared with  the  other  dimensions.* 

Let  b  =  external  radius  of  coils  in  cm., 
a  =  internal  radius  of  coils  in  cm., 
2  d  =  thickness  of  plate  in  cm., 
JJL  =  permeability  (assumed  constant  for  entire  section  of 

plate), 
co  =  2  7m, 


P 

N  =  number  of  turns  per  cm., 
S  =  number  of  plates, 
7  =  intensity  of  current  in  coil. 
The  flux  in  each  plate  is 


_  b  4 

m        0g€2^       -2^  ' 


*  The  formulae  for  the  toroidal  coil  are  taken  from  lecture  notes  by  Prof. 
M.  I.  Pupin  and  are  published  here  with  his  permission. 


40  FORMULAE  AND  TABLES  FOR  THE 

The  total  flux  in  the  core  is 


m 


V 

a  v 


€2md_|_€-2md  _  2  COS  2  md 


(79) 


For  direct  current,  that  is,  for  complete  penetration  of  the  mag- 
netic flux  in  the  core,  we  have 

b 

a 
and 


(80) 


t.-l-t/ 

$'      2  dm  V 


€2  md  i  c-2  md  _  2  COs  2  md 


/01\ 


When  md  is  greater  than  unity,  which  is  generally  true  in  the  case 
of  high  frequencies, 

l-sii-  (82) 

As  an  illustration  we  may  take  the  data  of  the  coil  used  by  Mr. 
Alexanderson  in  his  experiments  on  the  effect  of  frequency  on  the 
permeability  of  iron.* 

'd  =  0.0038  cm.,    b  =  2.55  cm.,    a  =  0.65  cm.,     s  =  10,    N  =  20, 
(n  =  2.25  X  103,     p  =  104,     assumed.) 


n. 

m. 

|,  (by  equation  82). 

20,000 

4.2  X102 

0.31 

40,000 

5.96X102 

0.22 

80,000 

8.4  X102 

0.16 

120,000 

10.3  X102 

0.13 

160,000 

11.9  X102 

0.11 

200,000 

13.3  X102 

0.10 

These  values  agree  very  closely  with  the  experimental  values 
obtained  by  Mr.  Alexanderson  as  seen  from  the  curves  plotted  in 
Fig.  10.  The  increase  in  resistance  of  the  coil  due  to  the  eddy 
current  losses  in  the  iron  core  is  given  by  the  formula 


_  2  s/JV2co ,       b  sinh  2  md  +  sin  2  md 
m  a  cosh  2  md  -f  cos  2  md 


(83) 


Using  the  same  data  given  in  the  above  example, 
d  =  0.0038  cm.,  6  =  2.55  cm.,  a  =  0.65  cm.,  s  =  10,  N  =  20, 

*  Magnetic   Properties  of  Iron  at  Frequencies  up  to  200,000    Cycles, 
P.A.I.E.E.,  Nov.,  1911. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     41 
we  have  for  frequency  20,000  cycles  per  second 

t  /2  X  104  X  2.25  X  103 
m  =  &  TT  \l 


104 
2md  =  3.1, 


420, 


sinh  2  md  +  sin  2  md  .      ,  , 

cosh  2  md  + cos  2  me*  m  l  Approximately, 


log,  | -log,  3.91 -1.36. 


14.000 
12,000 
10,000 
8,000 


|6,000 


4,000 
2,000 


Magnetic  Flux  Densities 

at 

Constant  Ampere  Turns,  and  Varying  Frequency 
Experimental  Values  Plotted,  Calculated  Values  Marked  in  Circles 


20,000     40,000    60,000     80,000    100,000   120.000    140,000    160,000  180.000  200.000 
Frequency  in  Cycles  /Second 

FIG.  10. 


Introducing  these  values  in  (83)  we  get 

P      2X10  X  2.25  X  103  X  400  X  2T  X  2  X  104 

420  X  HP  =  7'32  Ohms' 

That  is,  the  energy  losses  in  the  iron  in  this  particular  case  are 
such  as  would  be  produced  by  an  additional  resistance  of  7.3  ohms 
in  the  circuit. 


42 


FORMULAE  AND  TABLES  FOR  THE 


TABLE.  I 

VALUE  OF  m  FOR  COPPER  CONDUCTORS 

The  volume  resistivity  of  high  conductivity  annealed  copper  at  60°  F. 
(15.6°  C.)  is  1696.5, 
hence  for  n  =  1 

m  =  0.2157  Vn 


n. 

m. 

n. 

m. 

5 

0.4824 

1,000 

6.8221 

10 

0.6822 

2,000 

9.6477 

15 

0.8355 

5,000 

15.255 

20 

0.9648 

10,000 

21.573 

25 

1.0787 

25,000 

34.109 

30 

1.1816 

50,000 

48.240 

35 

1.2763 

75,000 

59.080 

40 

1.3644 

100,000 

68.221 

45 

1.4472 

150,000 

83.55 

50 

1.5255 

200,000 

96.48 

60 

1.6710 

300,000 

118.16 

75 

1.8683 

400,000 

136.44 

100 

2.1573 

500,000 

152.55 

200 

3.0509 

600,000 

167.10 

300 

3.7366 

750,000 

186.83 

400 

4.3147 

1,000,000 

215.73 

500 

4.8240 

2,000,000 

305.09 

CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     43 


TABLE  II 

(For  use  in  Formulae  (7)  and  (67)  to  (77)) 


X. 

*fc). 

S  (x). 

ifc). 

»Ct), 

r(x). 

ic«). 

li  (x). 

z. 

0.00 

1 

0000 

1 

0000 

0 

0000 

0 

0000 

1 

000 

0 

0000 

0 

5000 

0.00 

0.05 

1 

0000 

1 

0000 

0 

0003 

0 

0006 

1 

000 

0 

0003 

0 

5000 

0.05 

0.10 

1 

0000 

1 

0000 

0 

0013 

0 

0025 

1 

000 

0 

0013 

0 

5000 

0.10 

0.15 

1 

0000 

1 

0000 

0 

0028 

0 

0056 

1 

000 

0 

0028 

0 

5000 

0.15 

0.20 

1 

0000 

1 

0000 

0 

0050 

0 

0100 

1 

000 

0 

0050 

0 

5000 

0.20 

0.25 

0 

9999 

0 

9999 

0 

0078 

0 

0156 

1 

000 

0 

0078 

0 

5000 

0.25 

0.30 

0 

9999 

0 

9998 

0 

0113 

0 

0225 

1 

000 

0 

0113 

0 

5000 

0.30 

0.35 

0 

9998 

0 

9997 

0 

0153 

0 

0306 

1 

000 

0 

0153 

0 

5000 

0.35 

0.40 

0 

9997 

0 

9995 

0 

0200 

0 

0400 

1 

000 

0 

0200 

0 

5000 

0.40 

0.45 

0 

9995 

0 

9992 

0 

0253 

0 

0506 

1 

000 

0 

0253 

0 

5000 

0.45 

0.50 

0 

9992 

0 

9987 

0 

0312 

0 

0624 

1 

000 

0 

0312 

0 

4999 

0.50 

0.55 

0 

9988 

0 

9981 

0 

0378 

0 

0754 

1 

001 

0 

0378 

0 

4999 

0.55 

0.60 

0 

9983 

0 

9973 

0 

0450 

0 

0897 

1 

001 

0 

0450 

0 

4998 

0.60 

0.65 

0 

9977 

0 

9963 

0 

0527 

0 

1051 

1 

001 

0 

0528 

0 

4998 

0.65 

0.70 

0 

9969 

0 

9950 

0 

0611 

0 

1217 

1 

001 

0 

0612 

0 

4997 

0.70 

0.75 

0 

9959 

0 

9935 

0 

0701 

0 

1394 

1 

002 

0 

0703 

0 

4996 

0.75 

0.80 

0 

9947 

0 

9916 

0 

0798 

0 

1581 

1 

002 

0 

0799 

0 

4995 

0.80 

0.85 

0 

9933 

0 

9893 

0 

0900 

0 

1780 

1 

003 

0 

0902 

0 

4993 

0.85 

0.90 

0 

9916 

0 

9866 

0 

1007 

0 

1988 

1 

003 

0 

1011 

0 

4991 

0.90 

0.95 

0 

9896 

0 

9834 

0 

1121 

0 

2205 

1 

004 

0 

1126 

0 

4989 

0.95 

1.00 

0 

9873 

0 

9798 

0 

1240 

0 

2430 

1 

005 

0 

1247 

0 

4987 

1.00 

1.05 

0 

9846 

0 

9756 

0 

1365 

0 

2664 

1 

006 

0 

1374 

0 

4984 

1.05 

1.10 

0 

9816 

0 

9708 

0 

1495 

0 

2903 

1 

•  008 

0 

1507 

0 

4981 

1.10 

1.15 

0 

9781 

0 

9654 

0 

1631 

0 

3149 

1 

009 

0 

1646 

0 

4977 

1.15 

1.20 

0 

9742 

0 

9593 

0 

1771 

0 

3399 

1 

Oil 

0 

1790 

0 

4973 

1.20 

1.25 

0 

9699 

0 

9526 

0 

1917 

0 

3652 

1 

013 

0 

1941 

0 

4969 

1.25 

1.30 

0 

9651 

0 

9451 

0 

2067 

0 

3906 

1 

015 

0 

2097 

0 

4963 

1.30 

1.35 

0 

9598 

0 

9370 

0 

2221 

0 

4162 

1 

017 

0 

2259 

0 

4957 

1.35 

1.40 

0 

9541 

0 

9282 

0 

2379 

0 

4416 

1 

020 

0 

2426 

0 

4951 

1.40 

1.45 

0 

9478 

0 

9186 

0 

2541 

0 

4668 

1 

023 

0 

2598 

0 

4944 

1.45 

1.50 

0 

9410 

0 

9083 

0 

2706 

0 

4916 

1 

026 

p 

2776 

0 

4936 

1.50 

44 


FORMULAE  AND  TABLES  FOR  THE 


TABLE  II   (Continued) 


X. 

*(*). 

S  (z). 

*«*). 

»(*). 

r  (z). 

J(*). 

li  (*). 

X. 

.50 

0 

9410 

0 

9083 

0 

2706 

0 

4916 

1 

026 

0 

2776 

0 

4936 

1.50 

.55 

0 

9337 

0 

8973 

0 

2875 

0 

5159 

1 

029 

0 

2959 

0 

4927 

1.55 

.60 

0 

9258 

0 

8857 

0 

3046 

0 

5395 

1 

033 

0 

3147 

0 

4917 

1.60 

.65 

0 

9175 

0 

8734 

0 

3219 

0 

5623 

1 

037 

0 

3340 

0 

4907 

1.65 

.70 

0 

9087 

0 

8605 

0 

3394 

0 

5841 

1 

042 

0 

3537 

0 

4895 

1.70 

1.75 

0 

8995 

0 

8471 

0 

3571 

0 

6049 

1 

047 

0 

3738 

0 

4883 

1.75 

1.80 

0 

8898 

0 

8332 

0 

3748 

0 

6245 

1 

052 

0 

3944 

0 

4870 

1.80 

1.85 

0 

8797 

0 

8188 

0 

3926 

0 

6429 

1 

058 

0 

4154 

0 

4855 

1.85 

1.90 

0 

8692 

0 

8041 

0 

4104 

0 

6599 

1 

064 

0 

4368 

0 

4840 

1.90 

1.95 

0 

8583 

0 

7891 

0 

4281 

0 

6756 

1 

071 

0 

4585 

0 

4823 

1.95 

2.00 

0 

8472 

0 

7738 

0 

4457 

0 

6898 

1 

078 

0 

4806 

0 

4806 

2.00 

2.05 

0 

8357 

0 

7583 

0 

4632 

0 

7026 

I 

086 

0 

5029 

0 

4787 

2.05 

2.10 

0 

8241 

0 

7428 

0 

4805 

0 

7138 

1 

094 

0 

5256 

0 

4767 

2.10 

2.15 

0 

8122 

0 

7272 

0 

4976 

0 

7237 

1 

102 

0 

5485 

0 

4746 

2.15 

2.20 

0 

8002 

0 

7116 

0 

5144 

0 

7321 

1 

111 

0 

5716 

0 

4724 

2.20 

2.25 

0 

7881 

0 

6961 

0 

5309 

0 

7391 

1 

121 

0 

5949 

0 

4701 

2.25 

2.30 

0 

7759 

0 

6808 

0 

5470 

0 

7447 

1 

131 

0 

6185 

0 

4676 

2.30 

2.35 

0 

7637 

0 

6656 

0 

5627 

0 

7490 

1 

141 

0 

6421 

0 

4651 

2.35 

2.40 

0 

7515 

0 

6507 

0 

5780 

0 

7521 

1 

152 

0 

6659 

0 

4624 

2.40 

2.45 

0 

7393 

0 

6360 

0 

5928 

0 

7540 

1 

164 

0 

6897 

0 

4596 

2.45 

2.50 

0 

7272 

0 

6216 

0 

6072 

0 

7549 

1 

175 

0 

7137 

0 

4567 

2.50 

2.55 

0 

7152 

0 

6076 

0 

6210 

0 

7547 

1 

188 

0 

7376 

0 

4537 

2.55 

2.60 

0 

7034 

0 

5939 

0 

6343 

0 

7535 

1 

201 

0 

7616 

0 

4506 

2.60 

2.65 

0 

6917 

0 

5807 

0 

6471 

0 

7516 

1 

214 

0 

7855 

0 

4474 

2.65 

2.70 

0 

6801 

0 

5678 

0 

6594 

0 

7488 

1 

228 

0 

8094 

0 

4441 

2.70 

2.75 

0 

6687 

0 

5553 

0 

6711 

0 

7453 

1 

242 

0 

8333 

.0 

4407 

2.75 

2.80 

0 

6576 

0 

5432 

0 

6822 

0 

7412 

1 

256 

0 

8570 

0 

4373 

2.80 

2.85 

0 

6467 

0 

5316 

0 

6928 

0 

7366 

1 

271 

0 

8807 

0 

4337 

2.85 

2.90 

0 

6360 

0 

5203 

0 

7029 

0 

7314 

1 

286 

0 

9042 

0 

4301 

2.90 

2.95 

0 

6255 

0 

5095 

0 

7124 

0 

7259 

1 

302 

0 

9276 

0 

4264 

2.95 

3.00 

0 

6153 

0 

4990 

0 

7214 

0 

7199 

1 

318 

0 

9508 

0 

4226 

3.00 

CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     45 


TABLE  II   (Continued) 


X. 

*(z). 

S  (z). 

*(*>. 

w(x). 

r(x). 

I  (x). 

li  (x). 

z. 

3.00 

0 

6153 

0 

4990 

0 

7214 

0 

7199 

1 

318 

0 

951 

0 

4226 

3.00 

3.05 

0 

6053 

0 

4890 

0 

7298 

0 

7137 

1 

334 

0 

974 

0 

4188 

3.05 

3.10 

0 

5956 

0 

4793 

0 

7378 

0 

7072 

1 

351 

0 

997 

0 

4149 

3.10 

3.15 

0 

6862 

0 

4700 

0 

7452 

0 

7005 

1 

368 

1 

019 

0 

4109 

3.15 

3.20 

0 

5770 

0 

4611 

0 

7522 

0 

6937 

1 

385 

1 

042 

0 

4070 

3.20 

3.25 

0 

5680 

0 

4525 

0 

7588 

0 

6867 

1 

402 

1 

064 

0 

4030 

3.25 

3.30 

0 

5593 

0 

4443 

0 

7649 

0 

6797 

1 

420 

1 

086 

0 

3990 

3.30 

3.35 

0 

5409 

0 

4364 

0 

7707 

0 

6726 

1 

438 

1 

108 

0 

3949 

3.35 

3.40 

0 

5427 

0 

4288 

0 

7760 

0 

6654 

1 

456 

1 

130 

0 

3909 

3.40 

3.45 

0 

5348 

0 

4215 

0 

7810 

0 

6583 

1 

474 

1 

151 

0 

3868 

3.45 

3.50 

0 

5270 

0 

4144 

0 

7856 

0 

6512 

1 

492 

1 

172 

0 

3828 

3.50 

3.55 

0 

5195 

0 

4077 

0 

7900 

0 

6441 

1 

510 

1 

193 

0 

3787 

3.55 

3.60 

0 

5123 

0 

4012 

0 

7940 

0 

6371 

1 

529 

1 

214 

0 

3746 

3.60 

3.65 

0 

5052 

0 

3949 

0 

7978 

0 

6301 

1 

547 

1 

234 

0 

3707 

3.65 

3.70 

0 

4984 

0 

3889 

0 

8013 

0 

6233 

1 

566 

1 

255 

0 

3666 

3.70 

3.75 

0 

4917 

0 

3831 

0 

8045 

0 

6165 

1 

584 

1 

275 

0 

3626 

3.75 

3.80 

0 

4853 

0 

3775 

0 

8076 

0 

6098 

1 

603 

1 

295 

0 

3586 

3.80 

3.85 

0 

4790 

0 

3721 

0 

8105 

0 

6032 

1 

622 

1 

314 

0 

3547 

3.85 

3.90 

0 

4729 

0 

3669 

0 

8132 

0 

5968 

1 

641 

1 

334 

0 

3508 

3.90 

3.95 

0 

4670 

0 

3619 

0 

8157 

0 

5904 

1 

659 

1 

353 

0 

3470 

3.95 

4.00 

0 

4613 

0 

3570 

0 

8181 

0 

5842 

1 

678 

1 

373 

0 

3432 

4.00 

4.05 

0 

4557 

0 

3523 

0 

8203 

0 

5781 

1 

697 

1 

392 

0 

3394 

4.05 

4.10 

0 

4503 

0 

3478 

0 

8225 

0 

5721 

1 

715 

1 

411 

0 

3357 

4.10 

4.15 

0 

4450 

0 

3434 

0 

8245 

0 

5662 

1 

734 

1 

430 

0 

3310 

4.15 

4.20 

0 

4399 

0 

3391 

0 

8264 

0 

5605 

1 

752 

1 

448 

0 

3284 

4.20 

4.25 

0 

4349 

0 

3349 

0 

8283 

0 

5548 

1 

771 

1 

467 

0 

3248 

4.25 

4.30 

0 

4300 

0 

3309 

0 

8301 

0 

5493 

1 

789 

1 

485 

0 

3213 

4.30 

4.35 

0 

4253 

0 

3270 

0 

8318 

0 

5439 

1 

808 

1 

504 

0 

3179 

4.35 

4.40 

0 

4207 

0 

3231 

0 

8334 

0 

5386 

1 

826 

1 

522 

0 

3145 

4.40 

4.45 

0 

4162 

0 

3194 

0 

8350 

0 

5335 

1 

845 

1 

540 

0 

3111 

4.45 

4.50 

0 

4118 

0 

3158 

0 

8366 

0 

5284 

1 

863 

1 

558 

0 

3078 

4.50 

46 


FORMULAE  AND  TABLES  FOR  THE 


TABLE  II   (Concluded) 


X. 

*(*). 

S  (z). 

»(«). 

»(*). 

r  (z). 

I  to. 

h  (*). 

z. 

4.50 

0 

4118 

{ 

3158 

0 

8366 

0 

5284 

| 

863 

1 

558 

0 

3078 

4.50 

4.55 

0 

4075 

( 

3123 

0 

8381 

0 

5235 

] 

881 

1 

576 

0 

3046 

4.55 

4.60 

0 

4033 

0 

3089 

0 

8396 

0 

5186 

] 

899 

1 

594 

0 

3014 

4.60 

4.65 

0 

3992 

0 

3055 

0 

8410 

0 

5139 

] 

917 

1 

612 

0 

2983 

4.65 

4.70 

0 

3952 

0 

3022 

0 

8424 

0 

5092 

] 

935 

1 

630 

0 

2952 

4.70 

4.75 

0 

3913 

0 

2990 

0 

8438 

0 

5047 

1 

953 

1 

648 

0 

2922 

4.75 

4.80 

0 

3874 

0 

2959 

0 

8452 

0 

5002 

1 

971 

1 

666 

0 

2893 

4.80 

4.85 

0 

3837 

0 

2928 

0 

8465 

0 

4958 

1 

989 

1 

684 

0 

2864 

4.85 

4.90 

0 

3800 

0 

2899 

0 

8479 

0 

4915 

5 

007 

1 

702 

0 

2835 

4.90 

4.95 

0 

3764 

0 

2869 

0 

8492 

0 

4873 

025 

1 

720 

0 

2807 

4.95 

5.00 

0 

3729 

0 

2841 

0 

8505 

d 

4832 

043 

1 

737 

0 

2780 

5.00 

5.05 

0 

3695 

0 

2813 

0 

8518 

0 

4792 

060 

1 

755 

0 

2753 

5.05 

5.10 

0 

3661 

0 

2785 

0 

8531 

0 

4752 

078 

1 

773 

0 

2727 

5.10 

5.15 

0 

3628 

0 

2758 

f 

8544 

0 

4713 

096 

1 

791 

0 

2701 

5.15 

5.20 

0 

3595 

0 

2731 

c 

8557 

0 

4675 

114 

1 

808 

0 

2675 

5.20 

5.25 

0 

3563 

0 

2706 

0 

8569 

0 

4637 

131 

I 

826 

0 

2650 

5.25 

5.30 

0 

3532 

0 

2680 

0 

8582 

0 

4600 

149 

1 

844 

0 

2626 

5.30 

5.35 

0 

3501 

0 

2655 

0 

8594 

0 

4564 

166 

1 

862 

0 

2602 

5.35' 

5.40 

0 

3471 

0 

2631 

0 

8607 

0 

4528 

184 

1 

880 

0 

2578 

5.40 

5.45 

0 

3441 

0 

2606 

0 

8619 

0 

4493 

201 

1 

897 

0 

2555 

5.45 

5.50 

0 

3412 

0 

2583 

0 

8631 

0 

4458 

219 

1 

915 

0 

2532 

5.50 

5.55 

0 

3383 

0 

2559 

0 

8643 

0 

4424 

236 

1 

933 

0 

2510 

5.55 

5.60 

0 

3355 

0 

2537 

0 

8655 

0 

4391 

2 

254 

1 

951 

0 

2488 

5.60 

5.65 

0 

3327 

0 

2514 

0 

8667 

0 

4358 

2 

271 

1 

969 

0 

2467 

5.65 

5.70 

0 

3300 

0 

2492 

0 

8678 

0 

4325 

2 

289 

1 

986 

0 

2446 

5.70 

5.75 

0 

3273 

0 

2470 

0 

8690 

0 

4294 

2 

306 

2 

004 

0 

2425 

5.75 

5.80 

0 

3246 

0 

2449 

0 

8701 

0 

4262 

2 

324 

2 

022 

0 

2404 

5.80 

5.85 

0 

3220 

0 

2428 

0 

8713 

0 

4231 

2 

341 

2 

040 

0 

2384 

5.85 

5.90 

0 

3195 

0 

2407 

0 

8724 

0 

4200 

2 

359 

2 

058 

0 

2364 

5.90 

5.95 

0 

3170 

0 

2387 

0 

8735 

0 

4170 

2 

376 

2 

076 

0 

2345 

5.95 

6.00 

0 

3145 

0 

2367 

0 

8746 

0 

4141 

2 

394 

2 

093 

0 

2326 

6.00 

CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     47 


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48  FORMULAE  AND  TABLES  FOR  THE 


CHAPTER  II 
INDUCTANCE 

Inductance.  —  The  subject  of  inductance  may  be  conveniently 
divided  into  three  parts:  viz.,  self  and  mutual  inductance  of  air- 
core  coils,  self  and  mutual  inductance  of  linear  conductors,  and 
inductance  of  iron-core  coils.  In  the  case  of  air-core  coils  the  in- 
ductance depends  only  on  the  number  of  turns  and  the  geometri- 
cal configuration  of  the  coil.  For  the  same  length  of  wire  we  may 
have  as  many  different  inductances  as  there  are  different  shapes 
in  which  the  coils  can  be  wound.  Obviously  since  the  inductance 
depends  upon  the  shape  of  the  coil,  we  shall  require  a  large  num- 
ber of  formulae  to  cover  all  possible  cases.  A  collection  of  formulae 
for  self  and  mutual  inductance  of  air-core  coils  by  E.  B.  Rosa  and 
Louis  Cohen  has  been  published  in  the  Bulletin  of  the  Bureau  of 
Standards.*  We  have  attempted  to  give  there  a  complete  list  of 
all  available  formulae,  and  for  this  reason  in  many  cases  we  gave 
several  different  formulae  for  the  same  case.  This  may  be  desir- 
able in  extremely  accurate  work,  as  it  makes  it  possible  to  check 
the  calculations  by  different  formulae  and  thus  discover  any 
possible  error. 

We  shall  make  use  of  that  paper  .here  in  the  discussion  of  self 
and  mutual  inductance  of  air-core  coils,  but  we  shall  only  give 
those  formulae  which  are  considered  of  most  practical  value,  and 
for  a  more  exhaustive  treatment  of  this  subject  the  reader  may  be 
referred  to  that  paper.  We  also  give  here  an  extensive  list  of 
formulae  for  the  self  and  mutual  inductance  of  linear  conductors. 
On  account  of  the  practical  importance  of  linear  conductors  in  the 
study  of  transmission  line  problems,  the  subject  is  covered  quite 
fully. 

Mutual  Inductance  of  Two  Coaxial  Circles.  —  There  are  a 
large  number  of  formulae  available  for  the  calculation  of  mutual 
inductance  of  twp  coaxial  circles.  Some  are  applicable  only  for 

*  Bulletin  Bureau  of  Standards,  Vol.  5,  pp.  1-132. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      49 


r  =  Vc2  +  d2.       a  —  -r— • 
Aa 

The  formula  for  the  mutual  inductance  is 


short  distances  between  the  circles,  others  are  applicable  only  for 
large  distances  between  circles  while 
some  are  more  general.  There  is 
one  formula  which  is  exact  for  all  pos- 
sible cases,  but  that  formula  involves 
the  use  of  elliptic  integrals  which 
makes  the  computation  laborious. 

We    shall    give    here    only    three 
formulas  which  will  cover  all  practi- 
cal cases. 
Let  a  and  A  denote  the  radii  of  the 

two  circles, 
c  denote  difference  in  the   two 

radii,  c  =  A  —  a, 
d  denote   distance  between  the 

circles,  ^ ^ ^ 


35 


(128): 


a. 


3 


FIG.  11. 


1575 


loge 


31 

2048' 


247 


2  (128): 
7795 


"6(128)»a     8(128)3V 


This  expression  gives  very  accurate  results  for  values  of  d  almost 
as  great  as  the  radius  a.  When  great  accuracy  is  not  required  this 
formula  may  be  used  even  for  considerable  larger  values  of  d. 

A  more  general  formula  which  is  applicable  in  almost  all  possible 
cases  with  respect  to  size  and  distance  of  circles  is  the  following: 

M  =  47r  Via  -  4irg*  (1  +  e),  (2) 

where 


R!  =  V(A  +  a)2  +  d2,        R2  =  V(A  -  a)2  +  d?, 

e  is  usually  a  very  small  quantity  and  can  be  neglected  in  most 

cases. 


50  FORMULAE  AND  TABLES  FOR  THE 

When  the  circles  are  of  nearly  the  same  radius  and  the  distance 
between  them  is  small  compared  with  the  radius,  we  have  for  an 
approximate  value  of  the  mutual  inductance  the  following  simple 
expression: 

M  =  47ra(log<^-2).  (3) 

When  only  a  rough  approximation  is  desired  formula  (3)  may 
be  used  even  when  the  circles  are  at  some  distance  from  each  other. 
Example.  — 

a  =  20  cm.,        A  =  20  cm.,        d  =  1  cm. 

By  formula  (1), 


log.8  =  log,  160  =  5.0752, 


M  =  4?r  X  20{[1  +  0.00048  -  •  •  •  ]5.075  -  [2  +  0.00016  -•••]! 
=  773.08  cm.  =  0.000773  mh. 
By  formula  (3), 

M  =  80  TT  (5.075  -  2)  =  772.88  cm. 

For  such  small  distances  the  error  in  formula  (3)  is  only  one 
part  in  a  thousand. 
Example.  — 

a  =  20  cm.,        A  =  25  cm.,        d  =  10  cm. 
By  formula  (1), 


8500      ,  OT™* 

log*  -  =  log       ,  _    =  log  16  =  2.7726, 

r2       1 


M  =  47T  V600{[1  +  0.04687  -  0.00092  +  •  •  •  ]  2.7726 

-  [2  +  0.01562  -  0.00094  +  •  •  •  ]  j  =  248.815  cm. 
=  0.0002488  mh. 

By  formula  (2)  we  have 

Ei  =  V(45)2  +  100  =  46.098,    R2  =  11.180, 

k  =  0.7927,     Vk  =  0.8903, 

1  =  0.05803,     q  =  0.02901,     ^=0.07027,     e  =  0.0024, 
M  =  16  7T2  V500  X  0.07027  X  1.0024 
=  248.72  cm. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      51 

By  formula  (3), 

i       8  a  160       o  1  1  nn 

log,       ==loc          =  3.1192, 


M  =  80  TT  (3.1192  -  2)  =  281.28  cm. 

It  is  seen  that  in  this  case  for  d  =  ^,    formulae  (1)  and  (2)  give 

2 

practically  identical  results,  while  formula  (3)  is  in  error  by  about 
13  per  cent. 

Attraction  Between  Circular  Currents.  —  The  force  of  at- 
traction or  repulsion  between  two  coaxial  circular  currents  may  be 
expressed  as  the  rate  of  variation  of  the  mutual  inductance  'with 
respect  to  the  distance  between  the  circular  currents,  and  is  given 
by  the  following  expression: 


p  =  ^L  =  —^  l927rV(l  +  20g2+  225  q*  +  1840  g6  H ),  (4) 

9  2       VAa 

where  a  •• 


a  and  A  are  the  radii  of  the  two  circles  and  z  is  the  axial  distance 
between  them. 

To  facilitate  calculations  we  give  a  table  at  the  end  of  the  chapter 
(Table  IV,  p.  79)  which,  together  with  the  formula,  is  taken  from 

Prof.  Nagaoka's  paper,*  giving  the  logarithm  of • —  for 

z          oz 

values  of  q  from  0.020  to  0.150  which  are  sufficient  for  most 
practical  cases. 
Example.  — 

a  =  A  =  10  cm.,        z  =  10  cm., 


V  =  -^=  =  y  ~  =  0.4472,     V  k'  =  0.6687, 
=  1  ~  °-6687 

q  =  0.0992. 

*  H.  Nagaoka,  "Attraction  between  Two  Coaxial  Circular  Currents,"  Tokyo 
Sugaku  —  Buturigakkwai  Kizi,  2nd  Ser.,  Vol.  6,  No.  10. 


52 


FORMULAE  AND  TABLES  FOR  THE 


From  Table  IV  by  interpolation  we  get 

-  VAa  dM 

dz 

.  m~ 
therefore  - 


logiof- 


}  =  0.85027, 


=  7.055 


and 


7.055  =  7.055. 


Mutual  Inductance  of  Two  Coaxial  Coils.  —  Let  there  be  two 
coaxial  coils  of  mean  radii  A  and  a,  axial  breadth  of  coils,  61  and  62 
radial  depth,  Ci  and  fy,  and  distance  apart  of  their  mean  planes  d. 
Suppose  them  uniformly  wound  with  n\  and  n2  turns  of  wire. 
The  mutual  inductance  M0  between  the  two  central  turns  of  the 
coils  is  given  by  formulas  (1)  to  (3)  and  the  mutual  inductance  M 
of  the  two  coils  of  HI  and  n2  turns  is  then,  to  a  first  approximation, 

M  =  nirizMo.  (5) 

Formula  (5)  would  be  absolutely  correct  if  all  the  turns  were 
concentrated  in  the  space  occupied  by  the  central  turns,  but 
owing  to  the  appreciable  sections  of  the  coils  a  correction  factor 
must  be  applied.  The  correction  term  is  usually  small,  not  over 
one  per  cent  in  most  cases. 

In  the  case  of  square-sectioned  coils  we  can  obtain  an  accurate 
determination  of  M  by  (5)  if  in  calculating  M o  we  use  a  modified 

radius  r  instead  of  the  mean  radius 
a,  the  value  of  r  being  given  by  the 
expression 

'-'(1+5ni).     w 

where  6  is  the  side  of  the  square 
section. 

If   the   coil  has   a    rectangular 


FIG.  12. 


section  it  can  be  replaced  by  two  filaments  (Fig.  12)  the  distance 
apart  of  the  filaments  being  called  the  equivalent  breadth  or  the 
equivalent  depth  of  the  coil. 


62-c2 


12 


2 18  is  the  equivalent  breadth  of  A, 
2  5  is  the  equivalent  depth  of  B. 


(7) 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      53 

The  equivalent  radius  of  A  is  given  by  the  same  expression 
which  holds  for  a  square  coil,  viz., 


1  24 

In  the  coil  B  the  equivalent  filaments  have  the  radii  r  +  5  and 
r  —  d  respectively,  where 


=  afl 


24:  a2 

The  mutual  inductance  may  now  be  accurately  determined. 
If  each  has  a  square  section,  it  is  necessary  only  to  calculate  the 
mutual  inductance  of  the  two  equivalent  filaments  given  by  (6). 
For  coils  of  rectangular  sections  as  A}  B,  the  mutual  inductance 
is  the  sum  of  the  mutual  inductances  of  the  two  filaments  A  on 
the  two  filaments  on  B  and  MQ  in  (5)  is  the  mean  of  the  four 
inductances  Mi3,  Mu,  M23,  MU  where  MM  is  the  mutual  induc- 
tance of  filament  1  on  filament  3,  etc. 


.764 


II 


12.236  cm 

FIG.  13. 


Example.  — 

a  =  A  =  25  cm.,     bi  =  4,     Ci  =  1,     d  =  10, 


1.25,     2/3  =  2.236  cm., 


the  distance  apart  of  the  two  filaments  which  replace  the  coil. 
We  now  find  by  formula  (1)  or  (2)  the  mutual  inductance  of  the 
two  circles  1,  2,  on  the  two  circles  3,  4,  where  a  =  25.00167  and  d 
=  7.764,  10  and  12.236  cm.  respectively.     Thus, 
2  M18  =  215.00228  TT, 

MM=    90.31304  TT, 

M23  =  130.  14060  TT 

4  M~  435.45592  *•' 
M  =  108.8640  TT, 

MQ    =  107.4885  TT 

AM          1.3755*-' 


54  FORMULAE  AND  TABLES  FOR  THE 

AM  =  the  correction  for  sections  of  the  coils  whose  dimensions 
are  given  above.  The  correction  factor  is  of  the  order  of  one 
per  cent  and,  therefore,  unless  great  accuracy  is  desired,  formula 
(5)  will  be  suitable  for  the  calculation  of  the  mutual  inductance 
of  coils. 

Mutual  Inductance  of  Concentric  Coaxial  Solenoids.  —  If  the 
solenoids  are  of  equal  length  and  we  denote  by  A  and  a  the  radii 
of  the  outer  and  inner  solenoids  respectively,  and  by  HI  and  nz  the 
number  of  turns  of  wire  per  centimeter  on  the  single  layer  wind- 
ing of  the  outer  and  inner  solenoids,  we  have  for  the  mutual  in- 
ductance 

M  =  4:TrWnln2(l-2Aa),  (8) 

l-r+A        a2    L     A3\        a4    /I  .2  A5     5A7 
where   «=  - 


35  a6 


/I  _  8A_7      4A°  _  3A")\ 
\7       7r7  "   r9  r11   /  H 


2048  A6\7 

r  =  VA2  +  I2. 

Putting  M  =  M  o  —  AM,  M0  =  4  7r2a2nin2Z  is  the  mutual  in- 
ductance of  an  infinite  outer  solenoid  and  the  finite  inner  solenoid, 
while  AM  is  the  correction  due  to  the  ends.  The  expression  for  a 
is  rapidly  convergent  when  a  is  considerably  smaller  than  A. 
Equation  (8)  shows  that  the  mutual  inductance  is  proportional  to 
Z  —  2  Aa,  that  is,  the  length  I  must  be  reduced  by  Act  on  each  end. 
For  the  case  of  two  solenoids  each  having  more  than  one  layer, 
formula  (8)  may  be  used,  A  and  a  being  the  mean  radii,  and  HI 
and  HZ  the  total  number  of  turns  per  centimeter  in  all  the  layers. 

When  the  solenoids  are  very  long  in  comparison  with  the  radii, 

A3   A5 
formula  (8)  may  be  simplified  by  omitting  the  terms  -^-j  -^  ,  etc. 

The  expression  for  a  then  becomes 

1         a2  a4  5  a6 


a  =  ~  — 


2      16  A2      128  A4      2048  A6 
Example.  — 

a  =  5  cm.,    A  =  10  cm.,     I  =  200  cm. 

Substituting  these  values  in  (8)  for  a  we  have 
a  =  0.4875  -  0.0156  -  0.0005  +  •  •  • 

=  0.4714, 

M  =  47T2  X  25nitt2  (200  -  9.428)  =  19057. 
If        HI  =  n2  =  10  turns  per  centimeter, 
M  =  0.0188  henry. 


(9) 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      55 

For  two  concentric  coaxial  solenoids  of  which  the  inner  sole- 
noid is  considerably  shorter  than  the  other,  we  have  the  following 
formula: 

.  a2A2l         1 
" 


in  which 


Pl  =  Vk2  +  A2,    where   Zi  = 


P2  =  Vk2  +  A2,    where   k  = 


FIG.  14. 


I  —  lz  —  h  =  length  of  inner  solenoid, 
x  =  length  of  outer  solenoid, 
A  and  a  are  the  radii  of  the  solenoids. 

This  is  a  very  accurate  formula,  and  the  series  being  rapidly  con- 
vergent is  very  convenient  for  computation. 
Example.  — 

I  =  5  cm.,  x  =  30  cm.,  A  =  5  cm.,  a  =  4  cm.,  n\  —  10,  n2  =  40, 

Pi  =  13.463,     P2  =  18.200, 

M  =  47r2  X  16  X  400J4.737  +  0.0122  -  0.0007J  =  11.99897  cm. 
=  0.0012  henry. 

In  the  case  of  two  coaxial  coils  or  solenoids  of  the  same  diameter 
and  the  same  number  of  turns  per  cm.,  we  can  also  calculate  the 
mutual  inductance  by  the  aid  of  self-inductance  formulae. 


56  FORMULAE  AND  TABLES  FOR  THE 

Let  LI,  L2,  1/s  denote  the  self  inductances  of  the  three  coils,  1,  2 
and  3  respectively. 

L    =  total  inductance  of  the  three  coils  in  series, 

Lr   =  inductance  of  coils  1  and  2  in  series, 

L"  =  inductance  of  coils  2  and  3  in  series, 

L     =  L!  +  L2  +  L3  +  2  MB  +  2  M23  +  2  Mu, 

L'   =  L1+L2  +  2M12, 

L"  =  L2  +  L3  +  2  M23, 

L    -  L'  -  L"  =  -  L2  +  2  M13, 

,  ,        L  -f-  L2  —  L   —  L  /i-,\ 

and  Mi3=-         —  ^—          —  (11) 

The  mutual  inductance  between  coils  1  and  3  is  expressed  in  terms 
of  self  inductance.  For  the  evaluation  of  the  self  inductances  in 
(11)  see  formulae  (13)  and  (14). 

Self  Inductance  of  a    Single  Layer  Coil  or  Solenoid.  —  The 
following  approximate  formula  for  the  self  inductance  of  long 

solenoids  is  often  given: 

(12) 


where  a  is  the  mean  radius,  HI  is  the  number  of  turns  of  wire  per 
cm.  and  I  is  the  length,  supposed  great  in  comparison  with  a. 
There  is  a  considerable  error  in  this  formula  due  to  the  end  effects. 
Prof.  Nagaoka*  has  recently  published  a  very  convenient  and 
accurate  formula  for  the  calculation  of  the  self  inductance  of  single 
layer  coils  which  is  of  the  following  form: 

L  =  4ir*a*ni*lK.  (13) 

a,  wi  and  I  have  the  same  significance  as  in  formula  (12)  and  K 
is  a  correction  factor  for  the  end  effects,  and  is  proportional  to 

~  .     Prof.  Nagaoka  has  also  prepared  a  table,  which  is  reproduced 

L 

here  (Table  V,  p.  80),  giving  the  values  of  K  for  values  of  -y-  from 

0.01  to  10.     Formula  (13)  is  very  accurate  and,  with  the  aid  of 
the  table,  very  convenient  for  calculations. 

For  very  short  coils,  we  also  have  the  following  convenient 
formula: 

2  f    . 


Jour.  Coll.  Sci.,  Tokyo,  Art.  6,  pp.  18-33;   1909. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      57 

n  is  the  whole  number  of  turns  of  wire  on  the  coil,  and  a  as  before 
is  the  radius  measured  to  the  center  of  the  wire. 
Example.  — 

a  =  10,        I  =  5  cm.,        n  =  10,  -y  =  4. 
From  Table  V  we  find  K  =  0.3654. 

L  =  47T2  X  100  X  100  X  5  X  0.3654  =  721,300  cm.  =  0.7213  mh. 
Computing  the  same  example  by  (14)  we  have 

log«  ^  =  loge  16  =  2.7725,       ^-2  =  0.0078,     n  =  50, 


L  =  47T  X  10  X  2500  {  2.7725  -  0.5  +.0.0078  (2.7725  +  0.25)  } 
=  720,714  cm.  =  0.7207  mh., 

which  agrees  very  closely  with  the  result  obtained  by  formula  (13), 
formula  (13)  giving  the  more  accurate  result. 

For  shorter  coils  the  two  formulae  give  results  in  still  closer 
agreement. 

In  the  derivation  of  formulae  (13)  and  (14)  a  uniform  distri- 
bution of  current  over  the  entire  surface  of  the  cylinder  was 
assumed,  a  condition  which  could  be  realized  only  if  the  winding 
were  of  infinitely  thin  strip  and  completely  covered  the  solenoid. 
A  winding  of  insulated  wire  or  of  bare  wire  in  a  screw  thread  may 
have  a  greater  or  less  self  inductance  than  that  given  above  by  the 
current  sheet  formulae  according  to  the  ratio  of  the  diameter  of  the 
wire  to  the  pitch  of  the  winding.  Putting  L  for  the  actual  value 
.of  the  self  inductance  of  a  winding  and  L8  for  the  current  sheet 
value  given  by  (13)  or  (14),  then 

L  =  L8-  AL. 

The  correction  AL  is  given  by  the  following  expression: 

+B).  (15) 


a  is  the  radius,  n  is  the  whole  number  of  turns  and  A  and  B 
are  constants  given  in  Tables  VI  and  VII. 

The  correction  term  A  depends  on  the  diameter  d  of  bare  wire 

and  the  pitch,  that  is,  on  the  value  of  the  ratio  -=•• 


58 


FORMULAE  AND  TABLES  FOR  THE 


If  in  above  example  we  assume  d  =  0.08  cm.,  then  •_  =  -^-^ 

=  0.8.    From  tables  (VI,  p.  82)  and  (VII,  p.  83)  we  get 

A  =  0.3337,  B  =  0.3186, 

AL  =  47r  X  10  X  50  (0.337  +  0.3186) 

=  4098, 
L,  =  721,300, 
and         L  =  721,300  -  4098  =  0.7172  mh. 

The  correction  in  this  case  is  about  0.6  per 
cent. 

The  Self  Inductance  of  a  Circular  Coil  of 
Rectangular  Section.  —  If  the  coil  is  not  very 
long,  the  length  being  less  than  the  diameter, 
we  have  the  following  expression  for  the  self 
inductance  of  a  circular  coil  of  rectangular  section,  which  is  very 
accurate. 

8<z  ,     v         ,      (16) 


FIG.  15. 


a  =  mean  radius  of  coil,  b  and  c  breadth  and  depth  of  coil, 
n  =  total  number  of  turns. 

The  values  of  2/1  and  2/2  are  given  in  Table  VIII,  p.  83,  as  func- 

b       c 

tions  of  x  =  -  or  =- ,  that  is,  x  is  the  ratio  of  the  breadth  to  depth 

of  the  section  or  vice  versa,  being  always  less  than  unity. 
Example.  — 

a  =  10  cm.,     6  =  1  cm.,     c  =  1  cm.,    n  =  100. 
From  Table  VIII  we  find 


2/i  =  0.8483,     2/2  =  0.8162, 


loge 


8a 


80 


4.0354, 


L  =  47r  X  10  X  104  { (1+  Tnftnr)  4-0354  -  0.8483  +  0.0005 } 
=  47r  X  318,900  =  4.0078  mh. 

For  the  self  inductance  of  long  solenoids  of  several  layers,  we 
have  the  following  approximate  formula:* 


Louis  Cohen,  Electrical  World,  Vol.  50,  1907. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      59 

8^c  ^ 


+  (w  -  2)  022  +    .  .  .  ]  f  Vai2  +  Z2  ~  gai) 

)  j  ,  (17) 


to 


where  ao  is  the  mean  radius  of  the  solenoid,  ai,  02,  as  ...  am 
are  the  mean  radii  of  the  various  layers,  m  is  the  number  of  layers, 
d  a  is  the  distance  between  centers  of  any  two  consecutive  layers, 
and  n  is  the  number  of  turns  per  unit  length. 

For  long  solenoids,  where  the  length  is,  say,  four  times  the 
diameter,  we  can  neglect  the  last  term  in  equation  (17). 

This  formula  is  sufficiently  accurate  for  most  purposes;  it  will 
give  results  accurate  to  within  two  per  cent  even  for  short  sole- 
noids, where  the  length  is  only  twice  the  diameter. 
Example.  —  A  solenoid  of  length  I  —  50  cm.,  mean  radius  5  cm. 
and  depth  of  winding  0.4  cm.,  is  wound  with  four  layers  of  wire 
of  500  turns  each. 

The  constants  in   formula  (17)  have  the   following  numerical 
values  : 

ao  =  5?     8a  =  Q.l,     I  =  50,     m  =  4,     n  =  10, 


ai  =  4.75,  a2  =  4.85  .  .  .  V4a02  +  J2  =  50.99,    v       +  ia  =  50.225. 

Introducing  these  values  in  (17)  we  get 

L  =  167T2n2  (1144.3  +  3360  -  10.84  -  1.04) 
=  70.56  mh. 

Self  Inductance  of  Circular  Ring.  —  The  self  inductance  of  a 
ring  of  circular  section  is  given  by  the  following  formula: 

4,rajlofry  -1.75J,  (18) 

where  a  is  the  radius  of  the  circle  and  p  is  the  radius  of  its  cross 
section. 

Empirical  Formula.  —  Recently  Profs.  Brooks  and  Turner* 
have  obtained  an  empirical  formula  for  the  self  inductance  of 
coils,  which  is  applicable  to  a  fair  degree  of  approximation  for  the 
calculation  of  the  inductance  of  any  closely-wound  cylindrical 

*  Morgan  Brooks  and  H.  M.  Turner,  "Inductance  of  Coils,"  University  of 
Illinois  Bulletin,  Vol.  IX,  No.  10. 


60 


FORMULAE  AND  TABLES  FOR  THE 


coil,  long  or  short,  thick  or  thin,  from  the  long  solenoid  to  the 
single  turn  of  fine  wire. 
The  formula  is  as  follows: 


(19) 


T 


I    Axis 


a  =  the  mean  radius  of  the 
winding, 

b  =  the  axial  length  of  the  coil ; 
for  single  turns  use  di- 
ameter of  wire, 

c  =  the  thickness  of  the  wind- 
ing, 

R  =  the  outer    radius   of   the 

winding, 

N  =  total  number  of  turns  in 
winding, 


,  = 


lAR 


F"  =0.51ogio(lOO  + 


14  # 


2  b  +  3  c> 


4 


For  spaced  windings,  formula 
(19)  is  not  recommended,  espe- 
cially when  the  space  between  the 
turns  is  great  as  compared  with 
the  diameter  of  the  bare  con- 
ductor. 

For  long  coils,  we  may  for  a 
first  approximation  consider  F'  and  F"  separately  equal  to  unity, 
and  formula  (19)  reduces  to  its  first  term 


(20) 


FIG.  16. 


+  c  + 


Formula  (20)  should  be  employed  only  when  the  length  6  is 
at  least  twice  the  outside  diameter  2  R.  The  error  involved 
scarcely  exceeds  4  per  cent  with  a  multiple  layer  winding,  and 
is  within  2  per  cent  for  a  single  layer  solenoid,  becoming  more 
accurate  as  the  length  of  the  coil  increases. 

To  check  the  accuracy  of  formula  (19)  we  will  work  out  by  this 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      61 

formula  the  same  examples  as  those  given  in  illustration  of  for- 
mula (13),  (16)  and  (17). 
Example.  — 

a  =  10  cm.,        6=5,          c  =  diameter  of  wire  =  0.10, 
N  =  50,  R  =  10.05, 

50  +  1.2  +  20.1 
F  =50  +  1  +  14.07  -L°96' 


F"  =  0.5  loglo(lOO  +  ^)=  L0278, 


T      4  7r2X  100X2500      1.096X1.0278 

L=         -ISIS""  "TOS"  0.734  mh. 

By  formula  (13)  the  value  of  the  inductance  for  this  coil  is  0.7207 
mh.,  hence  the  error  in  formula  (19)  for  this  case  is  about  2  per 
cent. 
Example.  — 

a  =  10,    6  =  1,    c  =  1,    N  =  100,    R  =  10.5, 

10  +  12  +  21 
=  10  +  10  +  14.7  = 

F"  =  0.51oglo(lOO  +  14X510'5)=  1.0566, 

L  =  47r2><1QOX104  X  1.239  X  1.0566  =  4.13  mh. 
iz.o 

By  formula   (16)   the  value  of  the  inductance  for  this  coil  is 
L  =  4.008  mh.,  and  the  error  in  formula  (19)  for  this  case  is  3 
per  cent. 
Example.  — 

o  =  5,     6  =  50,     c  =  0.4,     N  =  2000,     R  =  5.2, 

F'  =  1.007,    F"  =  1, 


r  25X4X10°      1.007      _ 

50  +  0.4  +  5:2-  X^T    =  71.14  mh. 

By  formula  (17)  the  value  of  the  inductance  for  this  coil  is 

L  =  70.56  mh. 
which  agrees  with  the  above  within  0.8  per  cent. 

As  an  extreme  case  we  may  consider  that  of  a  single  turn. 

Let  a  =  10  cm.,     p  =  0.5  cm. 

loge—  =  loge  160  =  5.075, 

and  by  formula  (18)  we  get 

L  =  47r  10  J5.075  -  1.751  =  417.83  cm. 


62 


FORMULAE  AND  TABLES  FOR  THE 


Working  out  the  same  example  by  formula  (19)  we  find 

L  =  413  cm., 

the  error  being  only  1  per  cent. 

From  these  examples  it  is  evident  that  formula  (19)  can  be 
relied  upon  to  give  results  with  an  accuracy  of  from  1  to  3  per 
cent. 

Table  IX,  p.  84,  is  reproduced  from  Prof.  Brooks'  and  Turner's 
paper  giving  the  values  of  Ff  and  F"  for  various  coil  proportions. 
Self  Inductance  of  Rectangle.  — 
Let  a  =  length  of  rectangle, 
b  =  breadth  of  rectangle, 
p  =  radius  of  conductor, 
d  =  V  a2  +  b\ 
The  formula  for  the  self  inductance  is 


L  =  4  5  (a  +  6)  loge •  —  a  loge  (a  +  d) 


(21) 


For  a  rectangle  made  up  of  a  conductor  of  rectangular  section 


L  =  4     (a  +  6)  lo& 


-  a  log,  (« 


+  2  d  +  0.447  (a  +  0) 


For  a  =  6,  a  square, 
L  =  8 


°'2235 


0.726). 


(22) 


(23) 


Mutual  Inductance   of  Two  Equal  Parallel  Rectangles. — 

For  two  equal  parallel  rectangles  of  sides  a  and  6  and  distance 
apart  d  the  mutual  inductance,  which  is  the  sum  of  the  several 
mutual  inductances  of  parallel  sides,  is 


+  8  [ a2  +  62  + 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      63 

For  a  square,  where  a  =  b,  we  have 
M  = 


+  8 


a  +  V2a*  +  d* 

a2  +  d2  -  2  Va2  +  d2  +  d]. 


(25) 


FIG.  17. 

Self  Inductance  of  Toroidal  Coil.  —  When  the  section  of  the 
coil  is  rectangular,  and  it  is  wound  uniformly  with  a  single  layer 
of  n  turns,  the  inductance  is  given  by  the  following  expression : 

TZ 
>i* 


(26) 


-R 


~e 


FIG.  18. 

r2  and  r\  are  the  outer  and  inner  radii  and  h  is  the  axial  depth  of 
the  coil.  For  a  coil  of  circular  cross  section  we  have  for  the  in- 
ductance,* 

L  =  4  7m2  (R  -  VR  -  a2),  (27) 

*  "A  Treatise  on  Alternating  Current  Theory, "  by  Alex.  Russell,  Vol.  I,  p.  52. 


64 


FORMULAE  AND  TABLES  FOR  THE 


where  R  is  the  mean  radius  of  the  coil  and  a  is  the  radius  of  its  cross 
section. 

Inductance  of  Linear  Conductors.  —  In  problems  dealing 
with  the  transmission  of  electrical  energy  by  alternating  currents 
over  long  distances,  we  must  know  accurately  the  electrical  con- 
stants, inductance,  capacity  and  resistance  of  the  lines.  We  shall 
therefore  try  as  far  as  possible  to  give  here  a  complete  list  of  all 
available  formulae  for  the  inductance  of  linear  conductors  of  various 
arrangement  of  circuits. 

The  simplest  case  is  that  of  a  single  straight  cylindrical  wire, 
whose  inductance  is  * 


071 

2  I  <  loge 2  ( >  approximately. 

OL  4 


(28) 


I  =  length  of  wire  in  cm.  and  a  =  radius  of  wire  in  cm.  When 
the  permeability  of  the  wire  is  /*  and  that  of  the  medium  outside 
the  wire  is  unity,  formula  (28)  assumes  the  following  form: 


(29) 


The  mutual  inductance  between  two  parallel  wires  separated  by 
a  distance  d  is 

I  + 


M  = 


=  2 1  <  log€  -r  +  -j  —  1  [  approximately. 


(30) 


If  in  the  case  of  two  parallel  wires  one  is  the  return  circuit  of  the 
other,  then  the  effective  inductance  of  both  wires  is  LI  +L2  —  2  M  and 

the  effective  inductance  of  each  wire  separately  is  —  -  —  :  — 

Zi 

When  the  two  wires  are  of  the  same  diameter,  the  effective  in- 
ductance of  each  wire  is  L  —  M  .  Using  the  values  of  L  and  M 
from  (28)  and  (30)  we  get 


(31) 


*  E.  B.  Rosa  and  Louis  Cohen,  Bull  Bureau  of  Stand.,  Vol.  5,  pp.  1-132. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      65 

Usually  -j  is  very  small,  hence 

(32) 
If  the  wires  are  of  magnetic  permeability  /*, 

(33) 


The  total  inductance  of  the  circuit  including  the  to  and  fro  con- 
ductors is 


a  '4' 

Equations  (28)  to  (34)  give  the  inductances  in  centimeters.  To 
express  the  inductance  in  mh.  per  km.,  we  must  multiply  each  of  the 
formulae  by  the  numerical  factor  0.1.  To  express  the  inductances 
in  mh.  per  mile,  we  must  multiply  each  of  the  formulae  by  the 
numerical  factor  0.161. 

If  we  replace  the  Naperian  logarithms  by  the  common  loga- 
rithms, we  may  put  (34)  in  the  following  form: 

L  =  1.483  logio-  +  0.161  mh.  per  mile 

°d  (35) 

=  0.921  logio-  +  0.10  mh.  per  km., 

d  and  a  being  expressed  in  the  same  units,  inches  or  centimeters. 
Formulae  (28)  to  (35)  were  derived  on  the  assumption  of  uni- 
form current  distribution  over  the  entire  sectional  areas  of  con- 
ductors, which  is  only  strictly  true  for  continuous  currents.     For 
alternating  currents  formula  (34)  assumes  the  form 


a         -  W 


where  a  is  a  constant  the  value  of  which  depends  on  the  frequency. 
For  low  frequencies  we  can  take  a  equal  to  J  and  for  very  high 
frequencies  a  is  zero.  An  accurate  expression  for  a  has  been  given 
by  Professor  Pedersen,*  which  is  as  follows: 


a  = 


li  (ma),  (37) 


2X109 
*  P.  O.  Pedersen,  Jahrbuch  der  Drahttosen  Tekg.,  Vol.  4,  pp.  501-515. 


66  FORMULAE  AND  TABLES  FOR  THE 

where  n  =  frequency  in  cycles  per  second, 


p 

V  =  permeability, 
p  =  resistivity  in  C.G.S.  units, 
a  =  radius  of  wire  in  cm. 

The  values  of  l\  (ma)  for  values  of  ma  from  0  to  6  are  given  in 
Table  II,  p.  43,  Chap.  I.  For  values  of  ma  greater  than  6  we 
have  the  following  approximate  expression: 

7  ,      ,      1.4142      0.5303       0.75 
li  (ma)  =  ----  7  —  vr  ~  7  —  \r  (38) 

ma          (ma)3       (ma)4 

Example.  —  What  is  the  value  of  a  for  a  copper  conductor  0.1  cm. 
radius  and  frequency  20,000  ? 


30.52, 

ma  =  3.05. 

From  Table  II  we  find   h  (ma)  =  0.4188, 
hence  a  =  0.2094. 

In  Table  X,  p.  85,  we  give  the  values  of  L  calculated  by  formula 
(35)  for  various  sizes  of  B.  &  S.  wire,  and  different  distances  apart. 

Inductance  of  Three-phase  Transmission  Lines.  —  The  in- 
ductance of  one  wire  of  a  three-phase  circuit  arranged  on  the 
corners  of  an  equilateral  triangle,  is  the  same  as  in  the  case  of  a 
two-wire  circuit  with  the  wires  the  same  distance  apart. 

(39) 

where  d  =  distance  between  any  two  wires  and  a  =  radius  of  wire. 

To  get  the  inductive  drop  between  two  wires,  the  inductances 

are  combined  geometrically,  the  inductance  of  circuits  (1.2),  for 

example,  being 

(40) 


If  the  wires  forming  the  circuit  are  arranged  in  a  straight  line  we 
have  for  the  inductance  of  the  middle  wire 


4    ' 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      67 

while  the  inductance  of  the  outer  conductors,  assuming  equal 
load  distribution  on  all  three  phases,  is  as  follows: 

L2  =  L3  =  21  j  loge~  +  0.625  j  -  (41) 

To  equalize  the  inductances  of  all  three  phases  the  wires  are 
transposed,  each  wire  taking  the  center  pin  for  one-third  the  dis- 
tance. Arranged  in  this  way  the  inductance  of  each  wire  is 

(42) 


In  equations  (39)  to  (42)  the  length  of  the  lines  and  the  inductances 
are  expressed  in  centimeters.  Expressing  the  inductances  in  mh., 
km.  or  mile,  and  at  the  same  time  convert  the  Naperian  loga- 
rithms into  common  logarithms,  equations  (40)  and  (42)  assume 
the  following  form: 

Lig  =  0.797  logio-  +  0.0866  mh.  per  km. 

°d  (40/) 

=  1.283  logio-  +  0.14  mh.  per  mile. 


Li2  =  0.797  logio-  +  0.166  mh.  per  km. 
=  1.283  logio-  +  0.27  mh.  per  mile. 


(42') 


Mutual  Inductance.*  —  Two  metallic  circuits  A,  B  and  C,  D 
suspended  on  the  same  pole  and  arranged  in  the  form  shown  in  Fig. 
(19).  The  mutual  inductance  between  A  and  CD  is 


the  mutual  inductance  between  B  and  CD  is 


Since  the  current  in  A  is  equal  to  that  in  B  but  of  opposite  sign, 
we  have  for  the  mutual  inductance  between  the  two  circuits 

-^l-  (43) 

tt    ) 

*  "  Theorie  der  Wechselstrome,"  von  J.  L.  La  Cour  and  O.  S.  Bragstad,  pp. 
547-555. 


68 


FORMULAE  AND  TABLES  FOR  THE 


If  the  circuit  CD  consists  of  only  one  conductor  C,  the  earth  being 
the  return  conductor,  then  we  may  put  to  a  high  degree  of  approxi- 
mation di  =  d3 

and  MAB_  c  —  2 1  loge  —•  (44) 


In  case  it  is  desired  to  avoid  inductive  effects  the  conductors  must 
be  so  arranged  as  to  make  did±  =  d^d^  so  that 


Such  an  arrangement  of  a  pair  of  circuits  is  shown  in  Fig.  20. 


FIG.  19. 


FIG.  20. 


Grounded  Conductors.*  —  The  self  inductance  of  a  linear  con- 
ductor suspended  at  height  h  above  ground,  and  the  ground  being 
the  return  conductor,  is 

***i* 

d  (45) 

=  0.74  logio  -  +  0.08  mh.  per  mile; 
a 

a  is  the  radius  and  I  is  the  length  of  conductor. 

The  mutual  inductance  between  two  grounded  parallel  wires  is 

+  0i  + 


+  fa  - 


-i 

.  per  mile, 


(46) 


*  O.  Heaviside,  Collected  Papers,  Vol.  1,  p.  101. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      69 


where  hi  and  hz  are  the  heights  above  ground  of  wires  1  and  2 
respectively  and  d  is  the  distance  between  them. 

When  the  wires  are  at  the  same  height  above  ground,  equation 
(45)  reduces  to  

M  =  0.74  log™  y rj~~  m^*  per  m^e>  (4^) 

Inductance  of  Split  Conductors.*  —  The  use  of  split  con- 
ductors has  been  proposed  for  long  distance  transmission  lines  as 
a  means  of  reducing  the  inductance  and  increasing  the  capacity 
of  the  line. 

Consider  first  the  case  of  a  looped  conductor  split  into  .two 
parts,  the  two  parts  of  each  conductor  being  placed  symmetrically 
with  respect  to  each  other,  and  the  distance  between  the  centers 
of  gravity,  D,  as  shown  in  Fig.  21.  Let  a  denote  the  radius  of 


Q 


FIG.  21. 

an  equivalent  conductor  whose  sectional  area  is  equal  to  the  sum 
of  the  sectional  areas  of  the  two  parts  of  the  split  conductor. 
Then  we  have  for  the  total  inductance  of  the  circuit  including 
the  to  and  fro  conductor 

L  =  2  Z  j  log,  ^  +  loge  ^  +  0.597  j  cm.  (48) 

L  =  0.322  j  log€  -  +  loge  ^  +  0.597  I  mh.  per  mile.       (49) 
(        a  a  ) 

d  =  5  a, 


or 


Example.  — 
Let 


a  =  0.1825  in.,  D  =  48  in., 
^  =  5.571,  loge  ^  =  3.962, 
L  =  3.26  mh.  per  mile. 

*  Louis  Cohen,  "Inductance  of  Split  Aerial  Conductors,"  Electrical  World, 
Nov.,  1912. 


70  FORMULAE  AND  TABLES  FOR  THE 

For  a  solid  conductor  of  the  same  radius  and  the  same  distance 
between  outgoing  and  return  conductor  we  find,  by  formula  (34), 

L  =  3.84  mh.  per  mile. 

Hence,  splitting  the  conductor  into  two  parts  and  separating  these 
parts  by  a  distance  equal  to  five  times  the  equivalent  radius  of 
solid  conductor  reduces  the  inductance  by  about  15  per  cent. 

Conductor  Split  into  Three  Parts.*  —  Let  the  three  parts  of 
each  conductor  be  arranged  symmetrically  at  the  vertices  of  an 


\ 


/  f-    -f 

,/ 

\ 


FIG.  22. 


equilateral  triangle  as  shown  in  Fig.  22.     The  inductance  of  the 
circuit  including  the  to  and  fro  conductor  is 


cm. 


0.322  \  |  loge  -  + 1  loge  ^  +  0.533  I  mh.  per  mile. 
/  o         a      o         d  ) 


(50) 


Considering  the  same  example  we  have  worked  out  for  the  case 
of  a  conductor  split  into  two  parts,  we  have 

D  =  48  in.,     a  =  0.1825  in.,     d  =  5  a 
and  L  =  3.07  mh.  per  mile. 

By  splitting  the  conductor  into  three  parts  and  separating  the 
parts  by  a  distance  equal  to  five  times  the  radius,  the  inductance 
of  the  circuit  is  reduced  by  about  20  per  cent. 

Self  Inductance  of  Two  Parallel  Hollow  Cylindrical  Con- 
ductors.f  —  Let  02,  cti,  and  62,  &i  denote  the  outer  and  inner 
radii  of  the  two  cylinders  respectively;  d  is  the  interaxial  dis- 
tance between  the  two  conductors,  and  I  the  length  of  each  con- 
ductor in  centimeters.  The  total  inductance  of  the  circuit  is 

02  ,  Ia22-3ai2 


of  -of 


*  1.  c. 

f  Alex.  Russell,  "Theory  of  Alternating  Currents,"  Vol.  I,  pp.  55-59. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      71 

If  we  put 

h2  =  022  -  ai2,     fc2  =  622  -  &!2, 
equation  (51)  can  be  written  in  the  following  form : 


,KO, 


To  reduce  (50)  and  (51)  so  as  to  give  the  inductance  in  mh.  per 
mile  it  is  necessary  to  multiply  the  right-hand  side  of  either 
equation  by  the  numerical  factor  0.161. 

Self  Inductance  of  Concentric  Conductors.  —  The  following 
formulae  apply  to  two  hollow  cylinders,  one  inclosed  in  the  other, 
and  insulated  from  it,  the  current  entering  one  cylinder  and 
returning  by  the  other.  Let  the  outer  and  inner  radii  of  the 
outer  and  inner  cylinders  be  denoted  by  62,  &i,  (h,  «i  and  let  I  be 
the  length  in  cms.  The  inductance  is 

02      1  a22-3a? 


cm- 


1   (622-6i2)2    &66i      2   &22- 

In  practice  the  cross-sectional  areas  of  the  two  cylinders  are 
always  equal,  that  is, 

and  formula  (53)  becomes 

2 


.  (54) 

If  the  inner  cylinder  were  solid,  ai  would  be  zero  and  622  would 
be  equal  to  6i2  +  «i2>  and  formula  (54)  would  reduce  to 


_ 


The  least  possible  value  of  L  is  when  61  =  a2  and  in  this  case 
L  =  I  (4  loge  2  -  2)  =  0.7726  Z. 


72  FORMULAE  AND  TABLES  FOR  THE 

For  very  high  frequencies  where  the  current  is  concentrated  along 
the  surface  of  the  conductor,  the  magnetic  force  is  confined  within 
the  space  between  the  conductors,  and  in  that  case  we  have  for  the 
inductance 

L  =  2Zlog£^  (56) 

Equations  (55)  and  (56)  may  be  combined  into  one  by  writing 
them  in  the  following  form: 


where  a  is  a  quantity  that  depends  on  the  frequency.     For  low 
frequencies  it  may  be  taken  equal  to  unity,  and  for  very  high 
frequencies  it  is  zero. 
Example.  — 

61  =  0.4  in.,        az  =  0.2  in., 


log,-1  =  0.693, 


-0578 


For  low  frequencies  a  =  1  and 

%  L  =  0.316  mh.  per  mile. 
For  very  high  frequencies  a  =  0  and 

L  =  0.223  mh.  per  mile. 

Triple  Concentric  Conductors.*  —  The  inductance  of  a  triple 
concentric  conductor  depends  on  the  current  distribution  in  the 
three  conductors.  Consider  three  hollow  cylinders  of  negligible 
thickness  and  radii  7*1,  r2,  r3,  beginning  with  the  inner  one.  Let 
the  current  7i  flow  through  the  inner  conductor  and  return  cur- 
rents 1  2  and  7s  through  the  middle  and  outer  conductors.  Then 


The  inductance  of  the  system  is 
L  = 


0.322  \  log,-  +  f^Y log,-  >  mh.  per  mile. 


(58) 


*  "Theory  of  Alternating  Currents,"  by  Alex.  Russell,  Vol.  I,  Chap.  2. 
See  also  "The  Magnetic  Circuit,"  by  V.  Karapetoff,  Chap.  XI. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS       73 


If  a  current  1  2  flows  through  the  middle  conductor  and  /i,  73  are 
the  return  currents  through  the  inner  and  outer  conductors, 

L  =  0.322  j  (j\*  loge^  4/rY  loge  ^  |  mh.  per  mile.       (59) 

If  a  current  73  flows  through  the  outer  conductor  and  the  return 
currents  7i,  72  through  the  inner  and  middle  conductors, 


L  =  0.322  \  (Y^  loge  -  +  loge  -  I  mh.  per  mile. 

(  V  3/  Pi  7*2  ) 


(60) 


The  Self  Inductance  of  a  Circuit  Formed  by  Three  Conductors 
Whose  Axes  Lie  Along  the  Edges  of  an  Equilateral  Prism.  — 

We  will  assume  that  the  three  conductors  have  equal  radii  a, 
the  interaxial  distance  between  any  two  conductors  being  d.     Let 
a  current  7i  flow  through  conductor  No.  1  and  the  return  currents 
72  and  73  through  conductors  No.  2  and  3  respectively. 
The  inductance  of  the  system  is 


L  = 


=  0.161  |  4  loge -+  1  I   jl  -y^3>mh.  per  mile. 


(61) 


The  inductance  has  a  maximum  value 


when  either  72  or  73  is  zero  and  it  has  a  minimum  value 


when  72  =  73  =  J  I\. 

Compound  Conductors.  —  A  compound 
conductor  consisting  of  a  steel  core  and  a 
concentric  coating  of  copper  is  now  being  used 
in  many  cases  for  transmission  purposes. 

The  inductance  of  a  linear  conductor  con- 
sists of  two  parts,  one  of  which  is  due  to  the 
magnetic  field  outside  of  the  conductor  and 
the  other  is  due  to  the  magnetic  field  within 
the  conductor.  We  may  call  these  the  external  and  internal 


FIG.  23. 


74 


FORMULAE  AND  TABLES  FOR  THE 


inductances,  respectively.     In  the  expression  for  the  inductance 
of  a  linear  conductor 


The  first  term  is  the  external  inductance,  and  the  second  term 
the  internal  inductance. 

In  compounding  the  conductor  only  the  internal  inductance 


is  affected  and  the  term       is  replaced  by  the  following  formula  : 


* 


where 


(62) 


«i  =  radius  of  core, 
02  =  radius  of  shell, 

71  =  current  in  core, 

72  =  current  in  shell, 

jLti  and  ^2  are  the  permeabilities  of  core  and  shell,  re- 
spectively. 


The  above  formula  is  somewhat  complicated  for  numerical 
computation,  and  Mr.  Fowle  has  supplied  a  short  table  in  his 
paper,  which  we  reproduce  here,  giving  the  values  of  L»  for  the 
standard  sizes  now  in  commercial  use. 


TABLE  (A) 

INTERNAL   INDUCTANCE  OF  COPPER-CLAD  STEEL  WIRES  WHEN  THE  CON- 
DUCTANCE RATIO  OF  STEEL  TO  COPPER  is  12  PER  CENT 


Values  of  n. 

Internal  inductance  in  C.  G.  S.  units. 

Per  cent 
20 
30 
40 
50 

Lt-  =  0.149  MI 
Li  =  0.0506  MI 
L{  =  0.0208  MI 
Lf  =  0.00925  MI 

+  0.0646  M2 
+  0.102M2 
+  0.147M2 
+  0.  193  M2 

*  The  above  formula  was  first  given  by  Prof.  A.  Vaschy  in  his  book,  "  Traite 
d'electricite  et  de  Magnetisme, "  Vol.  1,  p.  367  and  was  independently  derived 
by  Mr.  F.  F.  Fowle,  Electrical  World,  Dec.  29,  1910. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      75 


TABLE  (B) 

INTERNAL  INDUCTANCE  OF  ALUMINUM-STEEL  WIRES,  WHEN  THE  CONDUCT- 
ANCE RATIO  OF  STEEL  TO  ALUMINUM  is  19.35  PER  OINT 


Values  of  n. 

Internal  inductance  in  C.  G.  S.  units. 

Per  cent 
20 

30 
40 

L{  =  0.127  MI 
Li  =  0.0328  MI 
Li  =  0.00871  MI 

+  0.0990  M2 
+  0.1861M2 
+  0.2624  M2 

In  the  above  two  tables  n  is  the  conductance  ratio  of  the  com- 
pound wire  to  a  solid  copper  wire  of  equal  size. 

Series  or  Parallel  Arrangement  of  Inductance.  —  If  several 
inductances  are  connected  in  series  the  total  inductance  is  the 
sum  of  the  several  inductances. 

Lt  =  Li  +  L2  +  L3  +  -  •  •  +  Ln.  (63) 

For  several  inductances  in  parallel,  we  have  for  the  total  in- 
ductance, 

-TT7 r^ (64> 

-Lil/2L4  •  •  •  Ln-\-  •  -  • 

If  the  mutual  inductance  between  the  various  inductance  coils  is 
taken  into  account,  then  for  two  coils  in  series  we  have  for  the 
effective  inductance, 

T  T        I      T        I     r>  -\r  f&A\ 

Jut  —  Jui  -r~  L/2  -\-  Z  M }  (yQ) 

or  Lt  =  L,  +  L2  -  2  M. 

The  first  of  the  above  two  formulae  applies  when  the  magnetic 
fields  of  the  two  coils  are  aiding,  and  the  second  formula  applies 
when  the  magnetic  fields  are  opposing. 
For  two  coils  in  parallel  we  have 

r     T  71 T9. 

(65) 


For  more  than  two  inductances  in  parallel  the  formula  for  the 
total  inductance  is  generally  very  complicated  if  the  mutual  in- 
ductance between  the  various  branches  is  to  be  included.  For 
the  case,  however,  of  cylindrical  conductors  symmetrically  ar- 
ranged, formulae  have  been  derived  for  the  total  inductance  of 
several  wires  including  the  mutual  inductance  between  the  wires.* 

*  "Inductance  and  Capacity  of  Linear  Conductors,"  The  Electrician, 
Feb.  14  and  21,  1913. 


76  FORMULAE  AND  TABLES  FOR  THE 

Such  an  arrangement  of  circuits  is  being  used  for  instance  in  the 
case  of  horizontal  antennae  which  generally  consist  of  a  number 
of  wires  hi  parallel  suspended  at  some  height  above  ground. 

Inductance  of  Wires  in  Parallel.  —  The  following  formulae 
(66)  to  (72)  were  derived  on  the  assumption  that  all  the  wires 
are  suspended  at  the  same  height  above  ground,  the  earth  being 
the  return  conductor;  also  that  all  the  wires  are  of  the  same  size 
and  equally  spaced. 

For  two  wires  in  parallel 

'  (66) 


where  £  is  the  total  inductance,  L  is  the  inductance  of  each 
individual  wire  and  M  is  the  mutual  inductance  between  the 
wires.     The  values  of  L  and  M  can  be  obtained  by  formulae  (45) 
and  (48).  ^ 
Example.  — 

a,  radius  of  wire  =  0.08  in., 

d,  distance  between  wires  =  24  in., 

h,  height  above  ground  =  80  ft. 

By  (45)  and  (48)  we  have 

0A      2h  .   1\      OJ       /160X12  .      _  \ 
L  =  2(\o&-+^=  2  log.  (-^g-  +  0.25) 

=  19.15  cm.  per  cm. 
'    .      4/i2      .      4X6400 
M=  loge-3r  =  1°&  --  o  --  =         cm*  per  Cm' 


d2 

In  the  expression  for  M  we  neglected  T^  as  being  very  small  com- 
pared with  unity. 
We  have  by  (66) 

P      19.15  +  8.76 

£  =  -    — = —     -  =  13.15  cm.  per  cm. 

For  three  wires  in  parallel 

•C  =    orl"  J      A  n,1  '  (67) 


Mi  is  the  mutual  inductance  between  wires  1,  2  and  2,  3;  M2  is 
the  mutual  inductance  between  1  and  3. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      77 

Since  the  wires  are  equally  spaced,  we  may  put  M2  =  MI  —  log,,  4 
=  Mi—  1.386  and  formula  (67)  becomes 

_  L(L  +  Mi-1.386)-2M1« 
3L-3Mi- 1.386 

Using  the  same  values  for  L  and  M  as  in  the  previous  example 
we  find  the  value  of  the  inductance  for  three  wires. 
r  -  19-15  (19.15 +8.76 -1.386) -2  (8.76)*  _       . 
3  X  19.15  -  3  X  8.76  -  1.386 

For  four  wires  in  parallel 

r  _  (L  +  Mi)  (L  +  M3)  -  (M!  +  M2)2  ,     , 

4L-4M2-2M!  +  2 

where  MI  =  M&  =  M23  =  M34, 

M2  =  Mis  =  M24, 


HH-H-H 

6  6  6  cb 


234 
FIG.  24. 

Formula  (69)  may  be  put  in  the  form 

r  _  (L  +  MQ  (L  +  M!  -  2.2)  -  (2  MJ  -  1.386)2 
4L-4M1  +  1.15/ 

Taking  again  the  values  of  L  =  19.15  and  MI  =  8.76,  we  get  for 
the  inductance  of  four  wires  in  parallel, 

L  =  10.93  cm.  per  cm. 
For  five  wires  in  parallel, 
£  = 

-2M  iM,  - 


-M  2)  +Afi  (2M3 

(71) 

where  MI  =  Mi2  =  M23  =  M34  =  M46, 

M2  =  Mi3  =  M24  =  M35, 
M3  =  Mi3  =  M25, 
M4  =  M15. 


78  FORMULAE  AND  TABLES  FOR  THE 

Expressing  the  inductance  in  terms  of  L  and  MI  only,  as  in  the 
previous  cases,  formula  (71)  reduces  to 
JC  = 


-  1.86)  +  5  M?  -  1.87^-4.18 

(72) 
For  the  values  L  =  19.15     and    Ml  =  8.76 

r  _  7022.7  +  4899.4  -  8662.3  +  3038.8  -  949 
"  1833.6  -  1642  +  383.7  -  16.38  -  4.18 

5349.5     ft  „. 
=  ,_  .  Q   =9.64  cm.  per  cm. 

OO4.o 

If  we  have  more  than  five  wires  in  parallel,  which  is  generally  the 
case  for  antennae  of  large  stations,  we  can  obtain  an  approximate 
value  of  the  inductance  in  the  following  manner:  Suppose  we  have 
an  antenna  consisting  of  ten  wires,  determine  the  inductance  of 
five  wires  in  parallel  and  denote  it  by  I/,  then  assume  that  the 
antenna  consists  of  two  wires  separated  by  a  distance  5  d  (where 
d  is  the  distance  between  adjacent  wires)  and  the  inductance  of 
each  wire  is  Z/.  By  formula  (66),  the  effective  inductance  of  the 
ten  wires  in  parallel  is 

L'  +  M' 
£=     -2—  ' 

where  M'  is  the  mutual  inductance  between  two  wires  separated 
by  a  distance  5  d. 

Taking  the  data  from  the  previous  examples,  we  have  for  the 
inductance  of  five  wires  in  parallel 

L'  =  9.64  cm. 
By  (48)  we  have 


P     9.64  +  5.54 

hence  £  =  -  —  =  7.6  cm.  per  cm. 

z 

If  the  antenna  were  200  feet  long,  we  should  have 

£  =  7.6  X  200  X  30.5  =  0.046  mh. 

If  the  antenna  consisted  of  twelve  wires,  for  instance,  we  could 
break  it  up  into  three  groups  of  four  wires  each  and  proceed  in  the 
same  manner  as  above  using  formulae  (68)  and  (70). 

The  above  formulae  for  the  inductance  of  parallel  wires  are  very 
useful  in  the  determination  of  the  capacity  of  antennae,  as  will  be 
shown  in  Chapter  III. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      79 

TABLE  IV 
(For  use  in  Formula  4.) 


r-VAadJfl 

A 

f-T/AadM-]. 

A 

Ios4  ,  -d,J 

logl°l  *  aTJ 

0.020 
0.022 
0.024 
0.026 
0.028 

1.03365 
1  .  13786 
1.23314 
1.32091 
1.40231 

10421 
9528 
8775 
8140 

0.106 
0.107 
0.108 
0.109 
0.110 

0.93984 
0.95193 
0.96395 
0.97590 
0.98778 

1218 
1209 
1202 
1195 

1188 

0.030 
0.032 
0.034 
0.036 
0.038 

1.47823 
1.54938 
1.61635 
1.67963 
1.73962 

7592 
7115 
6697 
6328 
5999 

0.111 
0.112 
0.113 
0.114 
0.115 

0.99958 
1.01132 
1.02298 
1.03459 
1.04612 

1180 
1174 
1166 
1161 
1153 

0.040 
0.042 
0.044 
0.046 
0.048 

1.79668 
1.85108 
1.90309 
1.95292 
0.00077 

5706 
5440 
5201 
4983 

4785 

0.116 
0.117 
0.118 
0.119 
0.120 

1.05760 
1.06901 
.08036 
.09166 
.  10289 

1148 
1141 
1135 
1130 
1123 

0.050 
0.052 
0.054 
0.056 
0.058 

0.04681 
0.09118 
0.13401 
0.17542 
0.21551 

4604 
4437 
4283 
4141 
4009 

0.121 
0.122 
0.123 
0.124 
0.125 

.11407 
.  12520 
1  .  13727 
1.14729 
1.15826 

1118 
1113 
1107 
1102 
1097 

0.060 
0.062 
0.064 
0.066 
0.068 

0.25439 
0.29213 
0.32881 
0.36450 
0.39927 

3888 
3774 
3668 
3569 
3477 

0.126 
0.127 
0.128 
0.129 
0.130 

1.16917 
1  .  18004 
1  .  19086 
1.20164 
1.21237 

1091 
1087 
1082 
1078 
1073 

0.070 
0.072 
0.074 
0.076 
0.078 

0.43317 
0.46625 
0.49857 
0.53017 
0.56108 

3390 
3308 
3232 
3160 
3091 

0.131 
0.132 
0.133 
0.134 
0.135 

1.22305 
1.23369 
1.24429 
1.25484 
1.26536 

1068 
1064 
1060 
1055 
1052 

0.080 
0.082 
0.084 
0.086 
0.088 

0.59136 
0.62103 
0.65013 
0.67868 
0.70672 

3028 
2967 
2910 
2855 
2804 

0.136 
0.137 
0.138 
0.139 
0.140 

1.27584 
1.28627 
1.29667 
1.30703 
1.31736 

1048 
1043 
1040 
1036 
1033 

0.090 
0.092 
0.094 
0.096 
0.098 

0.73428 
0.76137 
0.78802 
0.81425 
0.84009 

2754 
2709 
2665 
2623 
2584 

0.141 
0.142 
0.143 
0.144 
0.145 

1.32765 
1.33790 
.34812 
.35831 
.36847 

1029 
1025 
1022 
1019 
1016 

0.100 
0.101 
0.102 
0.103 
0.104 

0.86555 
0.87814 
0.89065 
0.90307 
0.91540 

2546 
1259 
1251 
1242 
1233 

1  99fi 

0.146 
0.147 
0.148 
0.149 
0.150 

.37859 
.38869 
.39875 
1.40879 
1.41879 

1012 
1010 
1006 
1004 
1000 

0.105 

0.92766 

80 


FORMULAE  AND  TABLES  FOR  THE 


TABLE  V 
NAGAOKA'S  TABLE  OF  VALUES  OF  THE  END  CORRECTION  K  AS  FUNCTION 

T-,          DIAMETER 
OF  THE  RATIO  -T — 

LENGTH 

(For  use  in  Formula  13.) 


Diameter 

K. 

AI. 

A,. 

Diameter 

K. 

AI. 

A,. 

Length 

0.00 

1.000000 

-4231 

+24 

0.45 

0.833723 

-3160 

+21 

0.01 

1.995769 

-4207 

26 

0.46 

0.830563 

-3139 

22 

0.02 

1.991562 

-4181 

24 

0.47 

0.827424 

-3117 

21 

0.03 

1.987381 

-4157 

25 

0.48 

0.824307 

ouyo 

21 

0.04 

1.983224 

-4132 

25 

0.49 

0.821211 

-3075 

21 

0.05 

0.979092 

-4107 

+25 

0.50 

0.818136 

-3054 

+21 

0.06 

0.974985 

-4082 

26 

0.51 

0.815082 

-3033 

21 

0.07 

0.970903 

-4056 

24 

0.52 

0.812049 

-3012 

21 

0.08 

0.966847 

-4032 

24 

0.53 

0.809037 

-2991 

20 

0.09 

0.962815 

-4008 

26 

0.54 

0.806046 

-2971 

21 

0.10 

0.958807 

-3982 

+25 

0.55 

0.803075 

-2950 

+20 

0.11 

0.954825 

-3957 

24 

0.56 

0.800125 

-2930 

20 

0.12 

0.950868 

-3933 

23 

0.57 

0.797195 

-2910 

20 

0.13 

0.946935 

-3910 

26 

0.58 

0.794285 

-2890 

20 

0.14 

0.943025 

-3384 

27 

0.59 

0.791395 

-2870 

20 

0.15 

0.939141 

-3857 

+23 

0.60 

0.788525 

-2850 

+19 

0.16 

0.935284 

-3834 

23 

0.61 

0.785675 

-2831 

19 

0.17 

0.931450 

-3811 

26 

0.62 

0.782844 

-2812 

20 

0.18 

0.927639 

-3783 

24 

0.63 

0.780032 

-2792 

19 

0.19 

0.923854 

-5761 

24 

0.64 

0.777240 

-2773 

19 

0.20 

0.920093 

-3737 

+24 

0.65 

0.774467 

-2754 

+19 

0.21 

0.916356 

-3713 

24 

0.66 

0.771713 

-2735 

19 

0.22 

0.912643 

-3689 

25 

0.67 

0.768978 

-2716 

19 

0.23 

0.908954 

-3664 

23 

0.68 

0.766262 

-2697 

18 

0.24 

0.905290 

-3641 

25 

0.69 

0.763565 

-2679 

18 

0.25 

0.901649 

-3616 

+23 

0.70 

0.760886 

-2661 

+18 

0.26 

0.898033 

-3593 

24 

0.71 

0.758225 

-2643 

19 

0.27 

0.894440 

-3569 

23 

0.72 

0.755582 

-2624 

17 

0.28 

0.890871 

-3546 

24 

0.73 

0.752958 

-2607 

18 

0.29 

0.887325 

-3522 

24 

0.74 

0.750351 

-2589 

18 

0.30 

0.883803 

-3498 

+22 

0.75 

0.747762 

-2571 

+  17 

0.31 

0.880305 

-3476 

24 

0.76 

0.745191 

-2554 

17 

0.32 

0.876829 

-3452 

23 

0.77 

0.742637 

-2537 

18 

0.33 

0.873377 

-3429 

23 

0.78 

0.740100 

-2519 

17 

0.34 

0.869948 

-3406 

22 

0.79 

0.737581 

-2502 

16 

0.35 

0.866542 

-3384 

+24 

0.80 

0.735079 

-2486 

+19 

0.36 

0.863158 

-3360 

22 

0.81 

0.732593 

-2467 

16 

0.37 

0.859799 

-3338 

23 

0.82 

0.730126 

-2451 

16 

0.38 

0.856461 

-3315 

22 

0.83 

0.727675 

-2435 

16 

0.39 

0.853146 

-3293 

23 

0.84 

0.725240 

-2419 

17 

0.40 

0.849853 

-3270 

+22 

0.85 

0.722821 

-2402 

+  16 

0.41 

0.846583 

-3248 

23 

0.86 

0.720419 

-2386 

16 

0.42 

0.843335 

-3225 

21 

0.87 

0.718033 

-2370 

15 

0.43 

0.840110 

-3204 

21 

0.88 

0.715663 

-2355 

16 

0.44 

0.836906 

-3183 

23 

0.89 

0.713308 

-2339 

17 

CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      81 


TABLE  V  (Continued) 


Diameter 

K. 

A* 

A* 

Diameter 

K. 

AJ. 

A* 

A* 

Length 

Length  ' 

0.90 

0.710969 

-  2322 

+  14 

2.50 

0.471865 

-  9292 

+  405 

0.91 

0.708647 

-  2308 

16 

2.60 

0.462573 

-  8887 

378 

0.92 

0.706339 

-  2292 

15 

2.70 

0.453686 

-  8509 

355 

0.93 

0.704047 

-  2277 

16 

2.80 

0.445177 

-  8154 

330 

0.94 

0.701770 

-  2261 

14 

2.90 

0.437023 

-  7824 

312 

0.95 

0.699509 

-  2247 

+  15 

3.00 

0.429199 

-  7512 

+  293 

0.96 

0.697262 

-  2232 

15 

3.10 

0.421687 

-  7219 

275 

0.97 

0.695030 

-  2217 

15 

3.20 

0.414468 

-  6944 

260 

0.98 

0.692813 

-  2202 

14 

3.30 

0.407524 

-  6684 

245 

0.99 

0.690611 

-  2188 

14 

3.40 

0.400840 

-  6439 

230 

1.00 

0.688423 

-10726 

+344 

3.50 

0.394401 

-  6209 

+  220 

1.05 

0.677697 

-10382 

330 

3.60 

0.388192 

-  5989 

207 

1.10 

0.667315 

-10052 

316 

3.70 

0.382203 

-  5782 

195 

1.15 

0.657263 

-  9736 

303 

3.80 

0.376421 

-  5587 

186 

1.20 

0.647527 

-  9433 

290 

3.90 

0.370834 

-  5401 

174 

1.25 

0.638094 

-  9143 

+278 

4.00 

0.365433 

-  5227 

+  168 

1.30 

0.628951 

-  8865 

266 

4.10 

0.360206 

-  5059 

161 

1.35 

0.620086 

-  8599 

255 

4.20 

0.355147 

-  4898 

152 

1.40 

0.611487 

-  8343 

244 

4.30 

0.350249 

-  4746 

141 

1.45 

0.603144 

-  8099 

236 

4.40 

0.345503 

-  4605 

138 

1.50 

0.595045 

-  7863 

+224 

4.50 

0.340898 

-  4467 

+  134 

1.55 

0.587182 

-  7639 

215 

4.60 

0.336431 

-  4333 

125 

1.60 

0.579543 

-  7424 

208 

4.70 

0.332098 

-  4208 

118 

1.65 

0.572119 

-  7216 

198 

4.80 

0.327890 

-  4090 

115 

1.70 

0.564903 

-  7018 

190 

4.90 

0.323800 

-  3975 

102 

1.75 

0.557885 

-  6828 

+184 

5.00 

0.319825 

-18321 

+2227 

-397 

1.80 

0.551057 

-  6644 

176 

5.50 

0.301504 

-16094 

1830 

-306 

1.85 

0.544413 

-  6468 

170 

6.00 

0.285410 

-14264 

1524 

-241 

1.90 

0.537945 

-  6298 

161 

6.50 

0.271146 

-12740 

1283 

-193 

1.95 

0.531647 

-  6137 

154 

7.00 

0.258406 

-11457 

1090 

-153 

2.00 

0.525510 

-11809 

+580 

7.50 

0.246949 

-10367 

+  937 

-127 

2.10 

0.513701 

-11229 

539 

8.00 

0.236582 

-  9430 

810 

-104 

2.20 

0.502472 

-10690 

499 

8.50 

0.227152 

-  8620 

706 

-  86 

2.30 

0.491782 

-10191 

465 

9.00 

0.218532 

-  7914 

620 

2.40 

0.481591 

-  9726 

434 

9.50 

0.210618 

-  7294 

10.00 

0.203324 

82 


FORMULAE  AND  TABLES  FOR  THE 


TABLE  VI 

VALUES  OF    CORRECTION   TERM  A,   DEPENDING  ON  THE  RATIO  ^  OF  THE 
DIAMETERS  OF  BARE  AND  COVERED  WIRE  ON  THE  SINGLE  LAYER  COIL 
(For  use  in  Formula  15.) 


d 
D' 

A. 

AI. 

d 
D' 

A. 

AI. 

1.00 
0.99 
0.98 
0.97 
0.96 
0.95 
0.94 
0.93 
0.92 
0.91 
0.90 
0.89 
0.88 
0.87 
0.86 
0.85 
0.84 
0.83  ' 
0.82 
0.81 
0.80 

0.5568 
0.5468 
0.5367 
0.5264 
0.5160 
0.5055 
0.4949 
0.4842 
0.4734 
0.4625 
0.4515 
0.4403 
0.4290 
0.4176 
0.4060 
0.3943 
0.3825 
0.3705 
0.3584 
0.3461 
0.3337 

100 
101 
103 
104 
105 
106 
107 
108 
109 
110 
112 
113 
114 
116 
117 
118 
120 
121 
123 
124 

19ft 

0.70 
0.69 
0.68 
0.67 
0.66 
0.65 
0.64 
0.63 
0.62 
0.61 
0.60 
0.59 
0.58 
0.57 
0.56 
0.55 
0.54 
0.53 
0.52 
0.51 
0.50 

0.2001 
0.1857 
0.1711 
0.1563 
0.1413 
0.1261 
0.1106 
0.0949 
0.0789 
0.0626 
0.0460 
0.0292 
0.0121 
-0.0053 
-0.0230 
-0.0410 
-0.0594 
-0.0781 
-0.0971 
-0.1165 
-0.1363 

144 
146 
148 
150 
152 
155 
157 
160 
163 
166 
168 
171 
174 
177 
180 
184 
187 
190 
194 
198 

0  7Q 

0  3211 

0.78 
0.77 
0.76 
0.75 
0.74 
0.73 
0.72 
0.71 
0.70 

0.3084 
0.2955 
0.2824 
0.2691 
0.2557 
0.2421 
0.2283 
0.2143 
0.2001 

127 
129 
131 
133 
134 
136 
138 
140 
142 

0.50 
0.45 
0.40 
0.35 
0.30 
0.25 
0.20 
0.15 
0.10 

-0.1363 
-0.2416 
-0.3594 
-0.4928 
-0.6471 
-0.8294 
-1.0526 
-1.3403 
-1.7457 

1053 
1178 
1335 
1542 
1823 
2232 
2877 
4054 

CALC  ULA  TION  OF  ALTERNA  TING  C  URRENT  PROBLEMS      83 


TABLE  VII 

VALUES  OF  THE  CORRECTION  TERM  B,  DEPENDING  ON  THE  NUMBER  OF 
TURNS  OF  WIRE  ON  THE  SINGLE  LAYER  COIL 

(For  use  in  Formula  15.) 


Number  of 
turns. 

B. 

Number  of 
turns. 

B. 

1 

0.0000 

50 

0.3186 

2 

0.1137 

60 

0.3216 

3 

0.1663 

70 

0.3239 

4 

0.1973 

80 

0.3257 

5 

0.2180 

90 

0.3270 

6 

0.2329 

100 

0.3280 

7 

0.2443 

125 

0.3298 

8 

0.2532 

150 

0.3311 

9 

0.2604 

175 

0.3321 

10 

0.2664 

200 

0.3328 

15 

0.2857 

300 

0.3343 

20 

0.2974 

400 

0.3351 

25 

0.3042 

500 

0.3356 

30 

0.3083 

600 

0.3359 

35 

0.3119 

700 

0.3361 

40 

0.3148 

800 

0.3363 

45 

0.3169 

900 

0.3364 

50 

0.3186 

1000 

0.3365 

TABLE  VIII 
TABLE  OF  CONSTANTS  FOR  FORMULA  (16) 


•  b     c 
-  or  r  . 
c    b 

Vi- 

ft. 

b     c 
-  or  r. 
c    b 

Vi> 

1/2- 

0.00 

0.50000 

0.1250 

0.55 

0.80815 

0.3437 

0.05 

0.54899 

0.1269 

0.60 

0.81823 

0.3839 

0.10 

0.59243 

0.1325 

0.65 

0.82648 

0.4274 

0.15 

0.63102 

0.1418 

0.70 

0.83311 

0.4739 

0.20 

0.66520 

0.1548 

0.75 

0.83831 

0.5234 

0.25 

0.69532 

0.1714 

0.80 

0.84225 

0.5760 

0.30 

0.72172 

0.1916 

0.85 

0.84509 

0.6317 

0.35 

0.74469 

0.2152 

0.90 

0.84697 

0.6902 

0.40 

0.76454 

0.2423 

0.95 

0.84801 

0.7518 

0.45 

0.78154 

0.2728 

1.00 

0.84834 

0.8162 

0.50 

0.79600 

0.3066 

84 


FORMULAE  AND  TABLES  FOR  THE 


TABLE  IX 

SHAPE-FACTORS,  Ff  AND  F",  FOR  CERTAIN  COIL  PROPORTIONS 
(For  use  in  Formula  19.) 


a. 

6. 

c. 

R. 

F1. 

r  • 

F'  F". 

19 

200 

2 

20 

1.008 

1.001 

1.009 

15 

200 

10 

20 

1 

.015 

1.001 

1.016 

19 

100 

2 

20 

1.015 

1.003 

1.018 

15 

100 

10 

20 

1 

.028 

1.002 

1.030 

19 

50 

2 

20 

1.029 

1.006 

1.035 

15 

50 

10 

20 

1 

.051 

1.004 

1.055 

19 

20 

2 

20 

1.064 

1.013 

1.078 

15 

20 

10 

20 

1 

.097 

1.008 

1.106 

19 

12 

2 

20 

1.095 

1.019 

1.116 

15 

12 

10 

20 

1 

.129 

1.011 

1.141 

19 

8 

2 

20 

1.125 

1.026 

1.154 

15 

8 

10 

20 

1 

.154 

1.013 

1.169 

19 

5 

2 

20 

1.163 

1.035 

1.204 

15 

5 

10 

20 

1 

.182 

1.015 

1.200 

19 

4 

2 

20 

1.182 

1.040 

1.229 

15 

4 

10 

20 

1 

.190 

1.016 

1.209 

19 

3 

2 

20 

1.205 

1.045 

1.259 

15 

3 

10 

20 

1 

.203 

1.017 

• 

1.223 

19 

2 

2 

20 

1.235 

1.054 

1.301 

15 

2 

10 

20 

1 

.216 

1.017 

1.237 

19 

1 

2 

20 

1.277 

1.065 

1.360 

15 

1 

10 

20 

1 

.232 

1.018 

1.254  . 

19 

0.5 

2 

20 

1.302 

1.073 

1.397 

15 

0.5 

10 

20 

1 

.241 

1.018 

1.263 

19 

0.2 

2 

20 

1.320 

1.079 

1.424 

15 

0.2 

10 

20 

1 

.246 

1.019 

1.270 

19 

0.1 

2 

20 

1.326 

1.081 

1.433 

15 

0.1 

10 

20 

1 

.249 

1.019 

1.273 

SQUARE  WINDING-SECTION  COILS 

17.5     5 

5      20 

.172 

1.023 

1.199 

19      2 

2      20 

.235 

1.054 

1.301 

19.5     1 

1      20 

.292 

1.097 

1.418 

19.75    0.5 

0.5    20 

.342 

1.163 

1.561 

19.9     0.2 

0.2    20 

.387 

1.29 

1.79 

19.95    0.1 

0.1     20 

.407 

1.42 

2.00 

CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      85 

i 

TABLE  X 
INDUCTANCE  PER  MILE  OF  COMPLETE  METALLIC  CIRCUIT,  FORMULA  35 


Size, 
B.  &S. 

Diameter 
in  inches. 

Distance 
in  inches. 

Inductance 
in  mh. 

Size, 
B.  &S. 

Diameter 
in  inches. 

Distance 
in  inches. 

Inductance 
in  mh. 

0000 

0.46 

12 

2.707 

4 

0.204 

12 

3.231 

18 

2.969 

18 

3.492 

24 

3.154 

24 

3.678 

48 

3.600 

48 

4.124 

000 

0.41 

12 

2.782 

5 

0.182 

12 

3.305 

18 

3.043 

18 

3.566 

24 

3.228 

24 

3.751 

48 

3.675 

48 

4.198 

00 

0.365 

12 

2.857 

6 

0.162 

12 

3.380 

18 

3.118 

18 

3.641 

24 

3.303 

24 

3.826 

48 

3.750 

48 

4.273 

0 

0.325 

12 

2.930 

7 

0.144 

12 

3.456 

18 

3.193 

18 

3.717 

24 

3.378 

24 

4.349 

48 

3.824 

48 

4.349 

1 

0.289 

12 

3.007 

8 

0.128 

12 

3.531 

18 

3.268 

18 

3.792 

24 

'3.453 

24 

3.978 

48 

3.900 

48 

4.424 

2 

0.258 

12 

3.080 

9 

0.114 

12 

3.606 

18 

3.341 

18 

3.867 

24 

3.526 

24 

4.052 

48 

3.973 

48 

4.500 

3 

0.229 

12 

3.157 

10 

0.102 

12 

3.678 

18 

3.418 

18 

3.940 

24 

3.603 

24 

4.124 

48 

4.050 

48 

4.571 

86  FORMULAE  AND  TABLES  FOR  THE 


CHAPTER  III 
CAPACITY 

THE  capacity  of  a  conductor  is  defined  as  the  ratio  of  its  charge 
to  its  potential,  when  all  other  conductors  are  at  zero  potential. 

C=§  (1) 

or,  according  to  Maxwell,  the  capacity  of  a  conductor  is  its  charge 
when  its  own  potential  is  unity  and  that  of  all  the  other  conductors 
is  zero. 

The  capacity  between  two  conductors  is  the  ratio  of  the  charge 
on  either  conductor  to  the  difference  of  potential  between  .the 
conductors,  assuming  also  that  all  other  conductors  are  at  zero 
potential,  otherwise  the  capacity  is  considerably  influenced  to  an 
extent  depending  on  the  proximity  of  the  other  conductors. 

In  general  the  capacity  of  a  conductor  depends  on  the  geometri- 
cal configuration  of  its  own  surface,  its  distance  from  other  con- 
ductors and  the  material  medium  between  them. 

The  problem  of  determining  the  capacity  of  a  conductor,  or  a 
system  of  conductors,  offers  considerable  mathematical  difficul- 
ties, and  there  are  not  therefore  as  many  formulae  available  for 
the  calculation  of  capacities  as  in  the  case  of  inductances.  For 
linear  conductors,  however,  which  are  the  most  important  in 
practice,  a  sufficient  number  of  formulae  have  been  derived  to 
enable  us  to  compute  the  capacity  of  any  arrangement  of  circuits. 
Owing  to  the  extreme  importance  of  the  electrical  constants  of 
linear  conductors  in  problems  of  transmission  of  electrical  energy, 
we  shall  give  here  an  extensive  list  of  such  formulae. 

The  unit  of  capacity  is  the  capacity  of  a  condenser  in  which 
unit  charge  produces  a  unit  difference  of  potential  between  its 
conductors,  which  may  be  either  in  electrostatic  or  electromagnetic 
units.  The  ratio  of  the  electrostatic  to  the  electromagnetic  unit 
of  capacity  is  equal  to  F2  where  V  =  3  X  1010  cm.  per  sec.  (velocity 
of  light); 

hence  C«=-CB,  (2) 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      87 

The  practical  unit  of  capacity  is  the  farad.  It  is  the  capacity 
of  a  condenser  which  has  a  potential  difference  of  one  volt  when 
charged  with  one  coulomb. 

1  farad  ==  10~9  C.G.S.  units. 

1  microfarad  =  KH5  C.G.S.  units. 

In  practice  the  microfarad  is  generally  used. 

If  C  is  the  capacity  in  microfarads  and  Ce  the  capacity  in  electro- 
static units,  we  have 

1015  „  1 


c  — 

' 


Ce. 


Parallel  or  Series  Arrangement.  —  In  electrical  circuits  we 
frequently  use  a  number  of  condensers  which  may  be  either  con- 
nected in  parallel  or  in  series.  For  the  parallel  arrangement  the 
total  capacity  is  the  sum  of  the  capacities  of  the  several  condensers, 
that  is, 

C  =  d  +  C2  +  <73  +  -  -  -  +  Cn.  (3) 


FIG.  25. 

In  the  series  arrangement  the  resultant  capacity  is  given  by  the 
following  expression: 


or 


C~i  /~i  /~¥  fi       | 

20304 ...  On  -r 

The  capacity  of  three  condensers  in  series  for  instance  is 

C=7T7T 


(4) 


(5) 


The  total  capacity  of  several  condensers  connected  in  series  is 
always  less  than  the  capacity  of  the  smallest  condenser  in  the 
series. 


88  FORMULAE  AND  TABLES  FOR  THE 

Example.  —  Three  condensers  of  capacities  Ci,  Cz,  C$  connected  in 
series. 

Let  Ci  =  0.01  mf.,     C2  =  0.05  mf.,     C3  =  0.1  mf. 

0.01X0.05X0.1       _  0.00005  f 

0.005  +  0.001+0.0005       0.0065  " 

which  is  less  than  the  value  of  Ci,  that  of  the  smallest  condenser. 
Capacity  of  Parallel  Plates.  —  If  the  area  of  each  plate  is 
denoted  by  A  sq.  cm.  and  the  distance  between  them  by  h  cm., 
the  capacity  of  the  condenser  is 

A 


~TT 


4wh  9  X  105 
For  circular  plates  of  radius  r,  we  have 


(6) 


If  the  space  between  the  plates  is  filled  with  a  uniform  dielectric 
of  specific  inductive  capacity  K,  then 

AK          1 


Formulae  (6)  to  (8)  are  only  approximate;  they  were  derived  on 
the  assumption  that  the  distribution  of  the  electric  force  between 
the  plates  is  uniform  throughout  the  entire  area.  This,  how- 
ever, is  not  the  case;  the  lines  of  force  curve  around  the  edges, 
and  the  correction  due  to  this  effect  cannot  be  neglected  in 
accurate  calculations.  This  correction  factor  has  been  determined 
only  for  circular  plates,  and  the  corrected  formula  for  this  case 
is  as  follows:* 


(9) 

where  t  is  thickness  of  the  plates  and  h  is  the  distance  between 
them. 

For  very  thin  plates 

~          1          ,  /1A, 

mf- 


*  G.  Kirchoff,  Gesammelte  Abhandlungen,  p.  112,  "Zur  Theorie  des  Con- 
densators."  See  also  J.  A.  Fleming,  "The  Principles  of  Electric  Wave 
Telegraphy,"  p.  176. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      89 

Example.  —  Capacity  of  two-plate  circular  condenser  of  the  fol- 
lowing dimensions. 

r  =  10  cm.,  h  =  0.2  cm.,  t  =  0.05  cm.  air  dielectric,  K  =  1. 

By  formula  (7), 

C  =  13.9XlO-5mf. 
By  formula  (9), 

C  =  14.51  XlO-5mf. 
By  formula  (10), 

C  =  14.44  X  10~5  mf  . 

It  is  seen  that  in  this  particular  case  the  capacity  as  computed 
by  formula  (7)  is  in  error  by  about  4.5  per  cent.  The  smaller  the 
area  of  the  plates  and  the  greater  the  distance  between  them,  the 
larger  is  the  correction  term. 

For  a  multiple-plate  condenser  consisting  of  n  plates,  the 
capacity  is  (n  —  1)  times  the  expression  for  capacity  given  in 
formulae  (6)  to  (8). 

When  the  dielectric  of  a  plate  condenser  consists  of  several  layers 
of  different  specific  inductive  capacities,  we  have,  on  neglecting 
the  correction  factor, 


where  hi  and  Ki  denote  the  thickness  and  specific  inductive  capac- 
ity of  layer  i,  and  n  is  the  number  of  layers,  or 


I^^^^M^^^ 

FIG.  26. 

Example.  —  Same  constants  as  in  the  preceding  example. 
Two-plate  circular  condenser, 

r    =  10  cm.,      A  =  TT  X  100, 

hi  =  hz  =  h%  =  /i4  =  0.05  cm., 

h  (distance  between  plates)  =  0.2  cm., 

#1  =  1.5,     Ki  =  2,     K3  =  2.5,     #4  =  3. 


90  FORMULAE  AND  TABLES  FOR  THE 

C== ^XlOO l_ 

(0.05     0.05  ,0.05  .0.5}  9  X  105 

"I  1.5  "      2      "  2.5  ""  3   J 

Capacity  of  Concentric  Spheres. — 

C  =  ^r^r  9xiQ5  mf .  (13) 

r2  is  the  inner  radius  of  outside  sphere  and  r\  is  the  radius  of 
inner  sphere.     This  formula  is  exact,  no  correction  term  is  re- 
quired. 
When  r2  is  infinitely  large 


which  is  the  capacity  of  an  isolated  sphere  in  infinite  space. 

From  formula  (14)  it  is  obvious  that  a  sphere  will  have  to  be  of 
radius  9  X  105  cm.,  or  9  km.,  to  have  a  capacity  of  one  microfarad. 

When  the  centers  of  the  spheres  are  separated  by  a  small  dis- 
tance d,  the  capacity  is  given  by  the  following  expression  : 

rif2 


f  f     ~ 

(ri  -  r2)  (n*  -  r23) 

If  the  dielectric  between  the  spheres  is  made  up  of  several 
spherical  layers  of  different  specific  inductive  capacities,  the  ex- 
pression for  the  capacity  is  as  follows: 

c=--  —  -  -  -;  (16) 


T\  is  the  radius  of  the  sphere  which  separates  the  layer  i  from  the 
layer  i  +  1. 

Comparing  formulae  (13)  and  (15)  it  is  seen  that  displacing  the 
centers  of  the  spheres  reduces  the  capacity. 
Example.  — 

TI  =  30  cm.,  r2  =  28  cm. 

By  formula  (13), 

30X28       1       _  1          f 

30-289X105  9X105 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      91 

If  the  centers  of  the  spheres  are  separated  by  0.5  cm.,  that  is,  in 
formula  (16)  d  =  0.5,  we  get 

30  X  28  {    _      210    j 
0      30  -  28 1         10,096 ) 


Displacing  the  centers  of  the  spheres  by  0.5  cm.   reduces  the 
capacity  by  two  per  cent  approximately. 
Capacity  of  Disk  insulated  in  Free  Space.  — 

<>  =  *>  (I?) 

d  =  diameter  of  disk  in  centimeters. 

Capacity  Coefficients  of  Two  Spheres.*  —  If  we  apply  a  differ- 
ence of  potential  to  two  spherical  conductors  which  are  separated 
from  each  other,  the  relation  between  the  charges  and  potentials 
are  given  by  the  following  equation: 

qi  =  CiVi  +  CuV2       and    q2  =  C2V2  +  C^Vi,  (18) 

where  Ci,  C2  and  CM  are  the  capacity  coefficients  of  the  spheres. 

Maxwell  f  derived  general  formulae  for  the  capacity  coefficients 
Ci,  C2  and  C&  covering  the  cases  of  equal  and  unequal  radii  of  the 
two  spheres  with  any  distance  separation  between  them.  The 
general  formulae,  however,  as  given  by  Maxwell  are  somewhat 
complicated,  so  we  shall  give  here  only  the  simplified  formulae 
obtained  by  Russell,  which  are  applicable  only  for  spheres  of 
equal  radii. 

Let  the  radius  of  each  sphere  be  a  and  d  the  distance  between 
their  centers;  then, 

r  =\'?          1 

f<  sinh  (2s  +  1) a 

.'I!  (19) 

-Ci2  =  x  T;     1    > 

s^J  sinh  2  sa 
where 

sinha  =  -     and     4X2  =  d2-4a2.  (20) 

*  Alexander  Russell,  Journal  of  the  Institution  of  Electrical  Engineers,  Feb., 
1912. 

f  Maxwell,  Electricity  and  Magnetism,  Vol.  1,  p.  270,  §  173. 


92 


FORMULAE  AND  TABLES  FOR  THE 


Denoting  -  by  y,  formulae  (19)  may  be  put  in  the  following  form: 
a 

C1  =  1+I+2  +  5       15      49      166 

o  y2     y      i/6     2/8     yw     y12 


and 


also 


=   l+-o  + 


OUi  A 

'~dd~A 


+ 1 


dd 


(21) 


(22) 


Ci  is  the  capacity  of  the  sphere  when  all  neighboring  conductors 
are  at  zero  potential.  When  the  charges  on  the  two  spheres  are 
+  q  and  —  q  respectively,  and  V  is  the  difference  of  potential 

between  them,  then  -^  is  the  capacity  C  between  the  spheres  and 

this  is  the  capacity  generally  considered  in  engineering  problems. 
In  this  case,  if  V  and  —  V  are  the  potentials  of  the  two  spheres, 


(23) 


When  both  spheres  and  all  neighboring  conductors  are  at  the 
same  potential  V,  then 

q  =  C,V  +  CnV. 

±=Cl  +  Cu  =  C'.  (24) 

V 

When  y  is  not  less  than  3  the  values  of  C  and  C1  are  given  by  the 
following  formulas: 


2C=     3/0/-1)2      jg  +  1       ly 

C^=      y  (t/  +  I)2      _  y-  1  _  1 

a       2/(2/  +  l)2  +  l         2/2         2/8' 


(25) 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      93 


For  values  of  y  greater  than  8  we  have  the  following  formulae: 

(26) 


2C  ,   1 

"     1" 


___. 

a  y      y2      y*      y* 

C         Cf 
In  Table  XI,  p.  122,  are  given  the  calculated  values  of  -  and  — 

for  values  of  -  from  2  to  1000. 
a 

The  Laws  of  Attraction  and  Repulsion  Between  the  Spheres 
When  the  Potentials  are  Given.  —  If  we  denote  by  W  the 
electrostatic  energy  of  the  system,  we  have 

W  =  \  C,  (TV  +  TV)  +  C^FiFj  (27) 

and  the  force  acting  between  the  spheres  is 

F  =  W  =  ~  \A  (Vl*  +  V**}  +  BVlVz'  (28) 

A  and  B  are  given  by  equation  (22). 
In  practice  three  cases  may  arise: 

(1)  When  the  potentials  are  equal  and  opposite. 

(2)  When  they  are  equal. 

(3)  When  one  of  the  spheres  is  at  zero  potential. 

In  the  first  case  the  force  F  is  attractive  and  is  given  by 


In  the  second  case  the  force  Fr  is  repulsive. 

1     (1~y)2  I  i2^  /»« 

4  "         V  • 


/oi\ 


In  the  third  case  the  force  F"  is  attractive. 

* 


94  FORMULAE  AND  TABLES  FOR  THE 


When  y  is  less  than  2.001  we  have  the  following  very  approxi- 
mate formulae: 

aV*_ 

(32) 


F  = 


2  (d  -  2  a) 
F'  =  0.07386  72. 

For  values  of  -  greater  than  10,  the  following  formulas  give  a  four- 
figure  accuracy. 

F=""i_^  • 

(33) 


Capacity  of  Linear  Conductors.  —  For  the  transmission  of 
electrical  energy,  linear  conductors  are  employed  in  the  form  of 
overhead  wires,  concentric  mains  or  cables.  In  some  of  these 
cases  the  derivation  of  capacity  formulae  offers  considerable 
analytical  difficulties.  A  sufficient  number  of  formulae  have  how- 
ever been  derived  to  cover  nearly  all  cases  which  are  apt  to  arise 
in  practice. 

Overhead  Wires.*  —  We  shall  first  consider  the  case  of  over- 
head lines  where  the  ground  acts  as  the  return  conductor,  which 
is  generally  the  case  for  telegraph  lines. 

If  we  have  a  single  wire  of  radius  a  suspended  at  height  h  above 
ground  its  capacity  is 


2  log.— 

I  is  the  length  of  the  line  in  centimeters;  h  and  a  can  be  expressed 
in  any  units  so  long  as  they  are  both  expressed  in  the  same  units. 
We  may  write  formula  (34)  in  the  following  form  : 

0.0894  /0_, 

(7  =  -  ^-r  mf  .  per  mile.  (35) 


*  See  O.  Heaviside,  Collected  Papers,  Vol.  I,  pp.  42-46. 
F.  F.  Fowle,  "Calculation  of  Capacity  Coefficients  of  Wires,"  Electrical 
World,  Vol.  58,  Aug.,  1911. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      95 

Formulae  (34)  and  (35)  give  only  an  approximation,  the  degree  of 
approximation  being  better  the  greater  the  ratio  -•    For  large 

values    of  -  formula   (34)  gives  fairly  accurate  results.      Prof. 

Kennelly  *  has  obtained  an  exact  formula  for  this  case  which  is  as 
follows: 

I  1 


C  = 


mf., 


2  cosh"1 - 
a 


or  we  may  put  it, 

Example.  — 
By  (35), 
By  (37), 


„       0.0894      ,  M 

C  = T  mf .  per  mile. 

cosh"1  - 
a 


h  , 
-  =  5. 
a 


(36) 


(37) 


C  = 


=  0.0384  mf.  per  mile. 
=  0.0390  mf .  per  mile. 


The  difference  in  results  by  the  two  formulae  is  about  1.5  per 
cent.     For  greater  values  of  -  the  results  by  the  two  f ormulae  will 

be  more  nearly  equal. 

Formula  (35)  was  derived  on  the  assumption  that  the  con- 
ductor is  suspended  in  free  space  and  that  there  are  no  other  con- 
ductors near  by  to  influence  the  value  of  its  capacity.  This  is  a 
condition  which  is  never  realized  in  practice.  Generally  there 
are  a  number  of  wires  suspended  on  the  same  pole  and  the  capacity 
of  each  conductor  is  affected  by  the  presence  of  the  other  con- 
ductors. In  the  following  we  shall  give  some  formulas  for  the 
case  of  two  or  more  grounded  wires  suspended  on  the  same  pole. 


A.  E.  Kennelly,  Proc.  Am.  Phil  Soc.,  Vol.  48,  pp.  142-165,  1909. 


96 


FORMULAE  AND  TABLES  FOR  THE 


Two  Grounded  Wires.  —  Two  wires  1  and  2  are  suspended  at 
heights  hi  and  h2  above  ground  and  at  distance  d&  from  each 
other  (see  Fig.  27). 

2 


.  r-*r- 

K 

\                        \ 

j 

I 

,  v 

\ 

A 


FIG.  27. 


The  capacities  of  wires  1  and  2  and  the  mutual  capacity  be- 
tween them  are 


0.0894  log,—2 

«2_ 

D 
0.0894  loge^ 


mf .  per  mile, 


D 

0.0894  loge^-2 

«12 


mf .  per  mile, 


mf .  per  mile, 


n      A      2hi\  A      2h2\     A      VV 

D=(loge-— )    (log,—-  J-(loge-T-). 

\         «i  /   \         az  /      \        «i2/ 


(38) 


where 


ai  and  02  are  the  radii  of  wires  1  and  2  respectively. 

When  the  wires  are  of  the  same  radii,  di  =  0%  =  a,  and  at  the 
same  heights  above  ground,  hi  =  h2  =  h,  formulae  (38)  reduce  to 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS      97 

0.0894  log,— 

Ci  =  C-2  = ,  r—  mf .  per  mile, 


(39) 
0.0894  log, 

mf.  per  mile. 


Example.  —  Single  wire,  No.  8  B.  &  S.,  suspended  25  feet  above 
ground. 


a  =  0.0642  in.,     log,       =  log,  =  9.142. 

0  0&Q4 
C  =  ^j^  =  0.00978  mf  .  per  mile. 

Example.  —  Two  grounded  wires,  No.  8  B.  &  S.,  equal  heights 
above  ground  25  feet  and  distance  apart  1  foot. 
As  in  the  preceding  example  we  have 


=  9.142,       log-  3.912. 

0.0894X9.142 
C2  =  2  _  (3.912)2  =  °-0120  mf-  Per 


„  0.0894X3.912 

Cl2  =  (9.142)2  -  (3.912)2  = 

An  additional  wire  on  the  same  pole  increases  the  individual 
capacity  of  each  wire.  In  the  case  considered  in  above  examples, 
the  increase  in  capacity  is  about  23  per  cent. 

As  we  increase  the  number  of  wires  on  the  same  pole  the  in- 
dividual capacities  of  the  several  wires  are  continually  increased. 
We  shall  give  here  formulae  for  the  capacities  of  three  and  four 
wires.  Though  the  formulae  are  not  very  simple,  and  are  unwieldy 
for  numerical  computation,  yet  the  numerical  examples  worked 
out  here  will  give  some  idea  of  the  magnitude  of  the  increase 
in  capacities  as  the  number  of  wires  is  being  increased. 

Three  Grounded  Wires.  —  The  wires  are  at  equal  heights  h 
above  ground,  the  same  distances  d  apart  and  of  equal  radii  a.- 
Wires  1  and  3  are  symmetrically  situated  with  respect  to  wire 


98 


FORMULAE  AND  TABLES  FOR  THE 


2,  hence  their  capacities  are  equal  and  for  the  same  reason  the 
mutual  capacities  Ci2  and  C^  are  equal. 


„ 


0.0894  j  (log.**)'-  (log.  Vag 
(\        a  I      \  d 


D 


~   _ 

0.0894 


D 


. 
)  mf  .  per 

"mile, 

mf.  per 
mile, 


2)  (log.  V^p  -  loge  2» 
l\  d  &    a 


C13 


i 

loge 


D 


.  vxl 

-  loge  -  X  loge 

J  a          to 


D 


f 
mf  .  per 

mile, 


, 

mf.  per 
mile, 


(40) 


where 


Example.  —  Use  same  constants  as  in  preceding  examples. 
a  =  0.0642  in.,     h  =  25  feet,     d  =  1  ft. 


=  9.142,  log, 


3.912,     log, 


3.219. 


D  =  764.5  +  98.54  -  279.84  -  94.80  =  488.4. 
0.0894J  (9.142)2-  (3.912)2 


C3  = 


0.0125  mf.  per  mile. 


0.0894K9.142)2-  (3.22)2J 
C2  =  -     —       4884  —  -  -  —  =  0.0134  mf.  per  mile. 


C.  =  C-    - 


5'92S 


=  -  0.0043  mf.  per  mile. 


0.0894J  (3.912)^9.142  X3.912J 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    99 


Four  Grounded  Wires.  —  Suppose  we  have  four  grounded 
wires  placed  at  the  corner  of  a  rectangle  as  shown  in  Fig.  28, 
and  we  shall  assume  that  the  wires  are  of  equal  radii  r. 

Let 


log.      =  d, 


The  significance  of  the  subscripts  to  h  and  d  are  obvious  from 
the  figure. 

1  2 

-o- 


FIG.  28. 

The  capacities  of  the  separate  wires  and  the  mutual  capacities 
are  as  follows  : 

-  J2)  +  d  (cf  -  de)  -c(df-  ce}\  +D,] 
CZ  =  C*  =  \e  (a2  -  62)  +  d  (be  -  ad)  +  c  (bd  -ac) {  +D, 
-  Cu=   \d  (df  -ce)+c  (cf  -de)+b  (<?  -  f)\  -f-D, 

(41) 


-  Cs4  =  f  d  (6d  -  ac)  +  c  (6c  -  ad)  +/  (a2  -  62)  j  -4-D, 
D  =  2  { (a»  -  62)(e2  -/2)  +  (c2  -  d2)2  +2  (ac  -  6d)(ay  -  ec) 


100  FORMULAE  AND  TABLES  FOR  THE 

Example.  —  The  wires  are  of  No.  8  B.  &  S.     r  =  0.0642  in. 
The  height  of  the  lower  two  wires  is  25  ft.  above  ground. 
The  vertical  distance  between  wires  d^  =  1  ft. 
The  horizontal  distance  between  wires  d^  —  2  ft. 
We  have  then 

hn  =  52  ft.,    hu  =  V2708,     ha  =  51,     hu  =  V2605, 
hn  =  50,         7*34  =  V2504,     du  =  2,       du  =  1,      d14  =  A/5, 
a  =  9.182,   b  =  3.259,    c  =  3.932,  d  =  2.628,  e  =  9.142,  /  =  3.22. 
D  =  2  [73.65  X  73.17  +  (8.52)2  -  2  X  2.75  X  27*.45 

-  2  X  11.33  X  11.38]  =  7388, 

671.7-29.93+107.88  ..  1.61 

X  —-  =  0.0181  mf.  per  mile, 


„       673.16-29.80-108.1  w  1.61 
C3  =  C4  =  -  —  X  --  =  0.013  mf  .  per  mile, 


-72.2-44.7  +  238.5     1.61 

X  -7T-  =  0.0029  mf  .  per  mile, 


„       33.5  +  37.1  -  252  _,  1.61 
-C24=  -Ci3  =  -  -  X  --  =  0.0044  mf.  per  mile, 


-  89.5  +  104.5  +  22.4      1.61     n  Annn     , 
-  C«  =  -  CM  =  -  -  X  --  =  0.0009  mf  .  per  mile, 


-  72.3  -44.5+  237.1      1.61 

7388 


-     .    -.  .    v/   .         n  nnon     , 

~  ~  X  ~~  ^  °'0029  mf'    er  mile- 


Capacity  of  Condenser  formed  by  Two  Long  Parallel 
Cylinders.*  —  The  capacity  of  a  looped  conductor  may  be  ex- 
pressed either  as  the  capacity  per  unit  length  of  each  wire  or  the 
capacity  per  unit  length  of  loop,  the  capacity  of  the  latter  being 
one-half  that  of  the  former.  In  single-phase  transmission  lines 
there  is  not  much  advantage  in  favor  of  one  or  the  other,  but  in 
the  case  of  three-phase  transmission  lines  it  is  of  some  advantage 
to  express  the  capacity  per  unit  length  of  each  wire. 

*  Alex.  Russell,  "The  Theory  of  Alternating  Currents,"  Vol.  I,  Chapters 
IV  and  V. 

"Theorie  der  Wechselstrome,"  J.  L.  LaCour  und  O.  S.  Bragstad,  pp. 
588-602. 

H.  Fender  and  H.  S.  Osborne,  "The  Capacity  between  Two  Equal  Parallel 
Wires,"  Electrical  World,  Vol.  56,  pp.  667-670,  Sept.,  1910. 

A.  E.  Kennelly,  "Graphical  Representation  of  the  Electrostatic  Capacity 
between  Wires,"  Electrical  World,  Vol.  56,  pp.  1000-1002,  Oct.,  1910. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    101 

In  formulae  (42)  to  (48)  we  shall  employ  the  following  notations : 
a  =  radius  of  conductor, 
d  =  interaxial  distance, 
h  =  height  above  ground. 

When  the  interaxial  distance  between  wires  is  large  compared 
with  the  radius  of  the  wire,  we  have  for  the  capacity  per  unit 
length  of  conductor 

(42) 


and  the  capacity  per  unit  length  of  looped  conductor 

(43) 


Expressing  the  capacity  in  mf.  per  mile,  formulae  (42)  and  (43) 
assume  the  forms 

0.0894 


(44) 


When  the  interaxial  distance  is  not  very  large,  we  must  use  the 
more  accurate  formula 

„  0.0894 

C  = ; — - ff   t  -  mf .  per  mile  of  single  wire.    (45) 


^fc+V&i 

In  the  derivation  of  formula  (45)  the  influence  of  the  earth,  as  a 
near-by  conducting  plane,  was  neglected.  Its  effect  is  negligible 
where  the  wires  are  at  a  considerable  height  above  ground, 
which  is  usually  the  case  for  transmission  lines.  Some  cases, 
however,  may  arise  where  it  is  desirable  to  take  into  account  the 
influence  of  the  earth  and  in  that  case,  if  we  have  two  wires  in 
the  same  horizontal  plane  height  h  above  ground,  the  capacity  of 
each  wire  is 

„  0.0894 

C  = : . (46) 


mf .  per  mile  of  single  wire. 


102  FORMULAE  AND  TABLES  FOR  THE 

If  the  two  wires  are  placed  one  vertically  above  the  other,  the 
capacity  is  given  by 

0.0894 ,  ..      , 

mf.  per  mile  of 


single  wire.  (47) 

If  d  is  small  compared  with  h,  equation  (47)  may  be  written 

C  = ; .     '          r- mf .  per  mile,       (48) 

2d2 


where  Ti,  hi  formulae  (47)  and  (48),  is  the  height  of  the  lower  wire 
above  ground. 

The  capacity  of  the  conductors  is  a  little  less  when  they  are 
placed  one  above  the  other  than  when  they  are  placed  in  the  same 
horizontal  plane,  assuming  that  the  distance  between  the  wires 
is  the  same  in  each  case. 
Example.  — 

a  =  0.1  in.,  d  =  2  ft.,  h  =  5  ft., 


log.  (l  +  5)  =  10&  1-04  =  0.039,    log,  (l  -  (d+daA).)  =  ~  <>-°28. 


0  08Q4 
By  formula  (46),      C  =  ,  .0      ^non  =  0.01643  mf.  per  mile. 


By  formula  (47),      C  =  ,  .Q'OQ  =  O-O164^  mf-  P^  mile. 
o.4o  — 


It  is  seen  from  the  above  example  that  even  when  the  conductors 
are  brought  within  5  feet  from  the  earth  the  capacity  is  increased 
only  by  about  0.7  per  cent,  so  that  in  all  practical  cases  we  can 
neglect  the  earth's  influence  and  use  the  simple  formula  (44)  or 
(45). 

When  the  radii  of  the  wires  are  not  equal,  as  has  been  assumed 
in  the  above  formulae,  we  have 

0.0894 
C  =  -  -,  -  r-mf.  per  mile  single  wire,       (49) 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    103 

•tn                   n  n 

d      d\       G&2 

where  a  =  — 


di  and  02  are  the  radii  of  the  two  wires. 

The  capacities  of  looped  conductors  can  be  also  expressed  very 
conveniently  in  terms  of  anti-hyperbolic  functions,  as  has  been 
done  by  Prof.  Kennelly.* 

When  the  radii  of  the  wires  are  equal  we  have 

0.0894 

mf .  per  mile  of  single  wire.  (50) 


cosh"1 


This  formula  is  equivalent  to  (45). 
For  unequal  radii  we  have 


(51) 


mf.  per  mile  of  single  wire,  which  is  equivalent  to  (49). 
Formulae  (45)  and  (50)  are  equivalent  to  each  other  since 


—  Riifl   f»nsh~l  — 


and  both  formulae  are  exact. 

In  Table  XII,  p.  122,  the  values  of  log€  -  and  cosh"1  ^-  are 

d  2i  O/ 

given  for  values  of  jr— from  1.01  to  25. 

£  d 

Formula  (44)  gives  fairly  accurate  results  when  ^—  is  greater 

Z  d 

than  5,  increasing  in  degree  of  accuracy  as  the  value  of  ^— in- 

2  d 

creases.    When  ^—  is  small,  that  is,  when  the  wires   are  close 

z  d 

together,  the  results  by  formula  (44)  are  considerably  in  error. 
*  A.  E.  Kennelly,  Proc.  Am.  Phil  Socy.,  Vol.  48,  pp.  142-165,  1909. 


104  FORMULAE  AND  TABLES  FOR  THE 

Example.  — 

For  s-  =  10. 

2a 

By  formula  (44), 

C  =  ^Qg!^  =  0-0298  mf.  per  mile. 

By  formula  (45)  or  (50), 

C  =  O'QQQO  =  0.0299  mf.  per  mile. 
For  ^  =  1.5. 

By  formula  (44), 

C  =  ^T™-  =  0.0814  mf.  per  mile. 
By  formula  (45)  or  (50), 

C  =  n'gR7ri  =  0.1031  mf.  per  mile. 


The  error  in  formula  (44)  for  this  case  is  about  27  per  cent. 
Example.  —  For  wires  of  different  diameters, 

a\  =  0.5  cm.,        az  =  1  cm.,       d  =  10  cm. 

By  formula  (49), 

100  -  1  -  0.25 

-  =  98.75, 


log,  \a  +  V^=lj  =  5.275, 


C=  1;.  =  0.034  mf.  per  mile  single  wire. 

2   (o.ZtO) 


Working  out  the  same  example  by  formula  (50)  we  get 

-1^  =  cosh-1  ^  =  2.9932, 
l.U 

cosh-1  ^  =  2.924, 


2.924) 
The  results  by  formulae  (49)  and  (50)  agree  exactly. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    105 


Two  Metallic  Circuits  Suspended  on  the  Same  Pole.*  —  If 

we  have  two  circuits  suspended  on  the  same  pole,  the  capacity  of 
either  circuit  will  be  sensibly  influenced  by  the  proximity  of  the 
other  circuit  and  the  capacity  will  be  larger  than  that  given  by 
formula  (44).  Suppose  we  have  two  circuits  (1,  2)  and  (3,  4)  on 
the  same  pole,  as  shown  in  Fig.  29.  The  capacity  of  circuit 
(1,  2)  may  be  expressed  by  the  formula 


6X  0.179 


d5 ,       dQ      A       dsd2 

—  loge  —  —  I  loge  -p-p 

a     3  a      \       didj 


mf .  per  mile  of  loop  conductor.  (52) 


FIG.  29. 

If  circuit  (3,  4)  were  removed  to  an   infinite  distance  from 
circuit  (1,  2)  we  should  have-i^  =  1  and  log€-^j  =  0. 

Equation  (52)  would  reduce  to 

0.179 
C  = 7-  mf .  per  mile  of  loop  conductor, 


which  is  formula  (44). 

If  the  circuits  are  separated  by  a  distance  of  5  feet  or  more, 
the  capacity  is  modified  but  slightly  on  account  of  the  proximity 
of  the  other  circuit.  If,  however,  the  circuits  are  brought  close 
together,  the  capacity  may  be  affected  materially. 

*  Louis  Cohen,  London  Electrician,  Feb.  14,  1913. 


106  FORMULAE  AND  TABLES  FOR  THE 

Example.  — 

a  =  0.1  in.     d  =  2  ft.  d&  =  2  ft.     di  =  I  ft. 

d2  =  3  ft.        d3  =.2.24  ft.      d4  =  3.6  ft. 

loge|6  =  log£^  =  log,  240  =  5.480, 
5  =  loge^-  =  Ioge240  =  5.480. 


loge         =  log£  1.866  =  0.624. 

ttitt4 

C  =  4X5.48X5^48  -(0.624)'  =  °-°0824  mf  '  per  mile'  lo°P  wire' 

If  circuit  (3,  4)  were  removed,  the   capacity  of  circuit  (1,  2) 
would  be 

C  =  ^   '  ,  ^  =  0.00812  mf.  per  mile  of  loop  wire. 

The  presence  of  circuit  (3,  4)  has  increased  the  capacity  of  (1,  2) 
by  1.5  per  cent. 

Three-phase  Transmission  Lines.*  —  If  the  spacings  are 
symmetrical,  as  in  the  case  when  the  wires  are  placed  at  the 
vertices  of  an  equilateral  triangle,  and  the  voltages  on  the  lines 
are  balanced  and  equal,  the  capacity  for  each  wire  is  the  same 
as  for  a  single-phase  transmission  line, 

n      0.0894     , 

C  =  -  -T  mf  .  per  mile. 


The  value  of  C  as  given  above  is  the  capacity  between  each  wire 
and  neutral,  and 

d  —  distance  between  wires, 

a  =  radius  of  wire. 

Charging  current  per  wire  is 

EaC 


j 

= 


For  unsymmetrical  spacing,  the  calculations  of  the  capacities 
or  charging  current  in  a  three-phase  line  is  very  much  involved, 
and  is  not  of  any  great  practical  importance.  It  is  therefore 
omitted  here. 

*  F.  A.  C.  Perrine  and  F.  G.  Baum,  Trans.  Am.  Inst.  E.  E.,  May,  1900. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    107 

We  may  get  an  idea  of  the  error  introduced  by  using  formula 
(44)  for  unsymmetrical  spacings  by  determining  the  limiting 
values  of  the  capacities  as  given  by  that  formula. 

As  an  illustration  let  us  consider  the  case  of  the  three  wires 
arranged  in  a  horizontal  plane.  The  distance  between  wires  is  12 
feet,  and  the  conductors  being  No.  0  B.  &  S.,  the  radius  a  =  0.163 
inches,  taking  d  =  12  feet  we  have 


log,-  =6.78, 


C  = 


0.0132  mf.  per  mile. 


For    d  =  24ft.,        log.-  =  7.37, 


C  = 


=  0.0121  mf.  per  mile. 


\ 


FIG.  30. 

Prof.  Karapetoff*  states  that  Mr.  J.  G.  Pertsch,  Jr.,  found  that, 
with  certain  simplifying  assumptions,  and  when  the  wires  are 
transposed,  the  equivalent  spacing  for  the  inductance  and  capac- 
ity is  equal  to  the  geometric  mean  of  the  three  actual  spacings,  or 

deq    =    V  di2  d<&  C?13. 

*  For  a  detailed  discussion  of  the  charging  currents  in  a  three-phase  line 
with  unsymmetrical  spacings,  see  "The  Electric  Circuit"  by  V.  Karapetoff, 
p.  199;  and  an  article  by  G.  S.  Humphrey,  Electrical  World,  Vol.  58,  p.  1300, 
Nov.  25,  1911.  See  also  "Inductance  and  Capacity  of  Three-phase  Transmis- 
sion Lines,"  by  Louis  Cohen,  Electrician,  July  18,  1913. 


108  FORMULAE  AND  TABLES  FOR  THE 

In  the  above  example  for  instance, 


deq  =  ^12X12X24  =  15.12,    loge       =  loge-'  =  7.012, 

=  0.0128  mf.  per  mile. 


Capacity  of  Horizontal  Antennae.*  —  A  horizontal  antenna  is 
generally  made  up  of  a  number  of  wires  in  parallel  suspended  at 
a  considerable  height  above  ground.  To  obtain  an  expression  for 
the  capacity  of  such  a  network  of  conductors  is  rather  a  difficult 
matter.  We  may,  however,  obtain  an  approximate  value  of  the 
capacity  of  antennae  by  an  indirect  method. 

For  linear  conductors  we  have  the  following  relation  between 
the  inductance  and  capacity: 

=        ' 


LC'  LV2' 

where  L  is  the  inductance  in  cms.  per  cm.,  C  is  the  capacity  in 
cms.  per  cm.  expressed  in  electromagnetic  units  and  V  is  the  veloc- 
ity of  light  (V  =  3  X  1010  cms.  per  cm.).  The  relation  given  by 
equation  (54)  is  only  strictly  true  when  there  is  no  energy  loss  in 
the  conductor,  that  is  with  perfect  conductivity.  The  expression 
for  C  as  given  by  equation  (54)  gives  fairly  accurate  results  if  in 
the  evaluation  of  L  we  neglect  that  part  of  the  inductance  which 
is  due  to  the  magnetic  field  within  the  conductor,  and  consider 
only  the  inductance  which  is  due  to  the  external  magnetic  field 
of  the  conductor. 

In  Chapter  II  we  have  given  a  set  of  formulas,  (65)  to  (71),  for 
the  calculations  of  the  inductances  of  two,  three,  four  and  five 
wires  in  parallel,  and  we  have  also  indicated  a  method  for  the 
approximate  determination  of  the  inductance  of  any  number  of 
wires  in  parallel;  hence,  knowing  the  inductance  we  can  use  for- 
mula (54)  to  determine  the  capacity. 

As  an  illustration  let  us  consider  an  antenna  of  the  following 
constants  : 

a  (radius  of  wire)  ^=  0.08  in., 

d  distance  between  adjacent  wires  =  2  feet, 

h  height  above  ground  =  80  feet, 

10  wires  in  parallel  and  150  feet  long. 

*  Louis  Cohen,  Electrician,  Feb.  21,  1913. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    109 


The  self  inductance  of  a  single  wire,  neglecting  the  internal  mag- 
netic field  of  the  conductor,  is 


2h 


160  X  12 


=18.65. 


The  mutual  inductance  between  two  adjacent  wires  is 

4X6400 


,,      , 

M  =  to-      = 


8.76. 


By  formula  (70),  Chapter  II,  we  find  the  inductance  of  five 
wires  in  parallel, 

£'  =  9.51  cms.  per  cm. 

The  mutual  inductance  between  two  wires  placed  at  a  distance  of 
5  d  is 


M' 


5.54  cms., 


and  the  total  inductance  of  the  10  wires  in  parallel  is 

r     L'  +  M'      9.51  +  5.54 

£  =  —  ~  -  =  -  jr  -  =  7.52  cms.  per  cm. 


The  capacity  of  the  antenna  is 

1015 1  1015  X  150  X  30.5 


C  = 


9  X  1020  X  7.52        9  X  1020  X  7.52 


=  0.00068  mf. 


Concentric  Cylinders. — The  capacity  of  a  condenser  formed  by 
two  concentric  cylinders  is 
Kl 


210*2 

0.0894  K 


mf .  per  mile, 


(55) 


where 


r2  =  inside  radius  of  outer  cylinder, 
7*1  =  radius  of  inner  cylinder, 
K  =  specific  inductive  capacity. 

If  the  dielectric  between  the  cylinders  is  not  homogeneous,  but 


110  FORMULAE  AND  TABLES  FOR  THE 

consists  of  several  cylindrical  layers  of  different  specific  inductive 
capacities,  formula  (55)  assumes  the  form 

0.0894 


(56) 
0.0894 


Ti    .      1    ,         r2    .      1 
-  +  ^-loge-  + 

r0      A2        TI      A 

Example.  — 

ri,  radius  of  inner  cylinder  =  1  cm., 
r2,  inner  radius  of  outer  cylinder  =  2  cm. 

The  dielectric  consists  of  two  layers  each  one-half  cm.  in  radial 
thickness,  the  specific  inductive  capacities  being  5  and  3,  respec- 
tively. 

Introducing  these  values  in  formula  (56)  we  get 

0.0894  0.0894 

C  =  1.5      1  -  2   =  0.0811+0.0959  =  a505  mf  '  per  mile' 


A  uniform  dielectric  of  specific  inductive  capacity  3  or  5 
would  give,  in  the  above  example,  a  capacity  of  0.39  mf.  per  mile 
or  0.65  mf.  per  mile,  respectively. 

If  the  axes  of  the  cylinders  are  displaced  with  respect  to  each 
other  by  a  distance  d,  the  capacity  is  given  by 


=  -    . 

loge(/3  +  V>_l) 

r22  +  n2  -  d2 

where      e=--2^r 

If  d2  is  small  compared  with  r22  —  n2,  then  approximately, 

0.0894  K  ,-Q, 

mf  .  per  mile.  (58) 


Another  approximate  formula  which  has  been  derived  for  this 
case  is  as  follows: 


C  =  i  +  -    -   mf.  per  mile.  (59) 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    111 


By  comparing  formulae  (58)  and  (59)  with  (55)  it  is  obvious  from 
inspection  that  the  capacity  is  a  minimum  when  the  axes  of  the 
cylinders  coincide.  As  an  illustration  of  the  relative  accuracy 
of  formulae  (58)  and  (59)  the  computed  values  of  the  capacities 

by  formulae  (57),  (58)  and  (59)  for  -  =  4  and  different  values  of 
d  are  given  in  the  following  table:  (K  =  1.) 


rz- 

TI. 

d. 

Formula  57. 

Formula  58. 

Formula  59. 

4 

1 

0 

0.0645 

0.0645 

0.0645 

4 

1 

1 

0.0676 

0.0676 

0.6750 

4 

1 

2 

0.0837 

0.0829 

0.0773 

4 

1 

2.5 

0.1111 

0.1054 

0.0837 

The  results  by  formulae  (57)  and  (58)  agree  very  closely  for 
all  values  of  d,  while  formula  (59)  is  applicable  only  when  d  is 
small.  For  comparatively  large  values  of  d,  the  error  in  formula 
(59)  is  considerable. 

Capacities  of  Cables.*  —  In  the  discussion  of  capacities  of 
cables  it  will  be  convenient  to  introduce  the  term  "effective 
capacity,"  which  we  shall  designate  throughout  by  Ce.  By  the 
term  effective  capacity  we  mean  that  factor  which,  multiplied  by 
the  time  rate  of  variation  of  electromotive  force,  will  give  the 
value  of  the  charging  current,  that  is,  7  =  Ce  2  irnE.  In  the  case 
of  a  simple  condenser,  the  effective  capacity  is  merely  the  capacity 
of  the  condenser,  but  in  the  case  of  cables  the  effective  capacity 
varies  with  any  alteration  in  the  other  parts  of  the  circuits  as  for 
instance  when  the  lead  sheath  is  grounded  or  insulated,  or  any  of 
the  other  cores  are  grounded,  etc. 

In  engineering  problems  we  are  mostly  interested  in  knowing 
the  values  of  the  charging  currents  for  a  certain  e.m.f.  and  a 
certain  frequency,  hence  it  will  be  very  useful  to  obtain  formulae 
for  the  effective  capacity  of  any  cable  system. 

*  Alex.  Russell,  "Theory  of  Alternating  Currents,"  Vol.  I,  Chapters  IV 
andV. 

L.  Lichtenstein,  "Capacity  of  Cables,"  Elektrotech.  Zs.,  Vol.  25,  pp.  106- 
110  and  124-126,  Feb.,  1904. 

L.  Lichtenstein,  Beitrage  zur  Theorie  der  Kabel;  Dissertation  Koniglichen 
Technischen  Hochschule  in  Berlin,  1908. 


112 


FORMULAE  AND  TABLES  FOR  THE 


Concentric  Cable.  —  We  shall  consider  first  the  case  of  a  simple 
cable  having  a  lead  sheath  and  an  iron  sheath  and  insulating 
dielectrics  between  them. 
Let 

TI  =  the  radius  of  conductor, 
r2  =  the  inner  radius  of  lead  sheath, 
r%  =  the  outer  radius  of  lead  sheath, 
r4  =  the  inner  radius  of  iron  sheath. 


FIG.  32. 


KI  is  the  specific  inductive  capacity  of  insulating  material 
between  conductor  and  lead  sheath  and  Kz  is  the  ^specific  inductive 
capacity  of  insulating  material  between  lead  and  iron  sheath. 

If  the  lead  sheath  is  grounded  the  capacity  of  the  cable  is  simply 
that  of  a  cylindrical  condenser  formed  by  conductor  and  lead 
sheath,  that  is,  by  formula  (55) 


!  0.0894 


mf .  per  mile. 


(60) 


If,  however,  the  lead  sheath  is  insulated,  charges  are  induced  on 
the  outer  surface  of  the  lead  sheath  and  the  inner  surface  of  the 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    113 

iron  sheath  and  thus  another  condenser  is  formed  between  lead 
and  iron  sheaths  whose  capacity  is 

f.  per  mile.  (61) 


The  effective  capacity  of  the  cable  is  that  of  the  two  condensers 
in  series,  that  is, 


If  KI  =  K2  we  have 

„  #0.0894 

Ce  —    

#0.0894 

mf .  per  mile. 

,      7-2r4 

lOge  — 

The  capacity  of  the  cable  is  reduced  when  the  lead  sheath  is  in- 
sulated. 
Example.  — 

ri  =  1  cm.,       7*1  =  1.5  cm.,    r3  =  1.75  cm.,    r4  =  2.5  cm., 

loge  -  =  0.405,       loge  -  =  0.356,       log,  —  =  0.761. 

C  =  0.221  K  mf.  per  mile, 
C'  =  0.251  K  mf.  per  mile, 
Ce  =  0.118#  mf.  per  mile. 

In  this  case,  by  insulating  the  lead  sheath,  the  capacity  of  the 
cable  is  reduced  to  about  50  per  cent  of  its  value  when  the  lead 
sheath  is  grounded. 
Triple  Concentric  Cable.  — 

Let          ri  =  the  radius  of  inner  conductor, 

r2  =  the  inner  radius  of  second  conductor, 
r3  =  the  outer  radius  of  second  conductor, 
n  =  the  inner  radius  of  third  conductor, 
r5  =  the  outer  radius  of  third  conductor, 
r6  =  the  inner  radius  of  lead  sheath. 


114 


FORMULAE  AND  TABLES  FOR  THE 


We  have  in  this  case  three  cylindrical  condensers  formed  be- 
tween the  several  conductors,  whose  capacities  are, 

n       Ki  0.0894 

Ci  =  mf .  per  mile, 


„       K2  0.0894     . 

62  = mf .  per  mile, 


„       K3  0.0894     .  ., 

(73  = mf.  per  mile. 


FIG.  33. 

• 

We  shall  assume  that  the  three  conductors  are  supplied  from  a 
three-p'hase  generator  delivering  an  electromotive  force  of  pure 
sine  form,  the  stator  being  star-connected. 

If  the  neutral  point  is  grounded,  the  charging  currents  have 
the  values 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    115 


72  = 


/         ST\) 

73  =  «JEfx  C2  V3  sin  co£  —  C3  sin  (  coZ  +  —  )  >  , 
(  \  6  /) 

in  (  ut  +  -^- 


s  sn 


(64) 


The  charging  currents  consist  of  two  components  differing  in 
phase  from  each  other  and  we  cannot,  therefore,  speak  of  effective 
capacity  in  the  case  of  a  triple  concentric  cable.  The  effective 
values  of  the  charging  currents  may  be  obtained  graphically. 

If  the  stator  winding  is  insulated,  neutral  point  not  being 
grounded,  we  have  for  the  charging  currents 


i  V3 


cos 


J2  = 


(65) 


J3  =  -co 

In  this  case  also  the  charging  currents  can  be  best  obtained 
graphically. 

Capacity  of  Two-core  Cable.  —  The  effective  capacity  of  a 
two-core  cable  can  be  expressed  for  all  possible  arrangements  in 
terms  of  Ci  and  Ci2,  where  Ci  is  the  capacity  of  No.  1  conductor, 
when  No.  2  and  the  sheath  are  grounded  and  Ci2  is  the  mutual 
capacity  between  the  two  conductors.  The  values  of  Ci  and  d2 
in  terms  of  the  dimensions  of  the  cable  are, 


log. 


0.0894  K 


/.      R*-d*y     L       d*  +  R*\ 
(log<-flH-(log<-2d/r) 


-Ca- 


-  per  mile, 


mf .  per  mile, 


(66) 


where 


R  is  internal  radius  of  sheath, 
r  is  the  radius  of  either  conductor, 
d  is  the  distance  between  centers  of  sheath  and 
either  conductor. 


116  FORMULAE  AND  TABLES  FOR  THE 

It  is  assumed  that  both  conductors  are  of  the  same  radius  and  are 
at  the  same  distance  from  center  of  sheath. 

If  the  sheath  is  grounded  and  the  two  conductors  are  con- 
nected to  a  generator  supplying  an  alternating  e.m.f.,  we  shall 
have  for  the  charging  current 


and  Ce  (effective  capacity)  =  J  (Ci  -  C12).  (67) 


FIG.  34. 

If  we  introduce  the  values  of  Ci  and  Cm  from  (66)  we  shall 
have 

mf .  per  mile.  (68) 


(69) 


Let  us  suppose  now  that  conductor  2  and  the  sheath  are  both 
grounded.  In  this  case  the  capacity  of  No.  1  conductor  is  simply 
Ci  and  that  of  No.  2  conductor,  Ci2;  that  is  the  charging  currents 
on  the  two  conductors  and  sheath  are 

71  =  CiwE  cos  co£, 

72  =  CizuE  cos  ut, 

I8  =  _  (d  +  Ci2)  uE  cos  at. 

If  one  terminal  of  generator  is  connected  to  the  two  conductors 
in  parallel  and  the  other  terminal  to  the  sheath  which  is  grounded, 
the  effective  capacity  of  either  conductor  is 

Ce   =  Ci  +  Cl2 

and  the  total  charging  current  on  the  two  conductors  is 

7  =  2  (Ci  +  Ci2)  uE  cos  at.  (70) 

Evidently  the  charging  current  on  sheath  is 

7.  =  -  7  =  -  2  (Ci  +  Ci2)  wE  cosorf. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    117 

In  all  the  above  cases  we  assumed  that  the  lead  sheath  was 
grounded.  If  the  lead  sheath  is  insulated,  and  one  terminal  of 
alternator  is  connected  to  the  two  conductors  in  parallel  and  the 
other  to  ground,  we  have  for  the  effective  capacity 

° 


where  Co  =  2  (Ci  +  CM) 

and  C  is  capacity  of  cylindrical  condenser  formed  between  lead 
and  iron  sheaths. 
Example.  — 

r  =  0.5  cm.,     d  —  1.0  cm.,     R  =  2.0  cm. 
(4  -1)X  0.0894  X 


i  = 


_ 
)  - 


t 

er  m 


_  Cl2  -  -  -  0.016  K  mf.  per  mile. 


For  the  case  given  by  formula  (68) 

<7e  =  0.051  K  mf.  per  mile. 
For  the  case  given  by  formula  (70) 

Ce  =  0.  134  Kmt.  per  mile. 

For  the  case  given  by  formula  (71)  we  must  also  determine  the 
value  of  C.  We  shall  assume  that  the  outer  radius  of  lead  sheath 
is  2.3  cm.  and  inner  radius  of  iron  sheath  is  2.6  cm.;  then, 

C  =  °'°89945  =  0.732  K  mf  .  per  mile, 


2  (Ci  +  Ci2)  =  0.134  K  mf.  per  mile 

C'C1 
and  Ce  =  —  ^fr  =  0.114  K  mf  .  per  mile. 

o  -f-  OQ 

Capacity  of  Three-core  Cables.  —  As  in  the  case  of  a  two-core 
cable  we  can  express  the  capacities  of  the  various  possible  ar- 
rangements in  terms  of  Ci  and  C&.  It  is  assumed  that  the 
conductors  are  symmetrically  placed  with  respect  to  the  axis  of 
the  sheath. 


118  FORMULAE  AND  TABLES  FOR  THE 

The  formulae  for  C\  and  Ci2  are 


an2  —  ai22 


an*   — 


2  ai2 


—  3 


(72) 


FIG.  35. 


where 

an  =  2  log* 


Rr 


R  =  radius  of  sheath, 
r  =  radius  of  conductor, 
d  =  distance  between  axes  of  conductor  and  axis  of  sheath. 

An  equivalent  formula  to  (72)  is  the  following: 
1  1 


61oge 


3  RWr 
1 


61oge 


-d1 


6  log. 


R*-d2 


(73) 


Formulae  (72)  and  (73)  were  both  derived  by  the  aid  of  the  prin- 
ciples of  images  and  give  identical  results. 

The  values  of  Ci  and  Ci2  in  formulae  (72)  and  (73)  are  given  in 
electrostatic  units.  To  convert  them  to  microfarads  per  mile  we 
must  multiply  by  the  numerical  factor  0.179  and,  since  the  con- 
ductors are  embedded  in  a  dielectric,  we  must  multiply  by  K  the 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    119 

specific  inductive  capacity  of  dielectric.      As  an  illustration  we 
shall  consider  a  cable  of  the  following  dimensions: 

r  —  0.5  cm.,     R  =  2.5  cm.,     d  =  1.3  cm. 
ai  =  2  log*  3.65  =  2.6, 
ecu  =  log*  1.66  =  0.51, 

=  j(2.6)2-  (Q.51)2jO.: 

1       (2.6)3  -  3  X  2.6  X  (0.51 

1  (0.51)2  -  2.6  X  0.51 1  K  X  0.179 
C12  -  (2.6)3_3X2>6X(0>51)2  +  2(0.51)2  =  -0.0119mf.permile. 

By  formula  (73)  we  find  for  the  same  example, 

7?2  _  ^72 

=  0.63, 


R*d* 

d  V3 


and 

Ci  =  0.074  K  mf  .  per  mile, 
CM  =  -  0.0119  K  mf.  per  mile. 

From  this  example  it  is  seen  that  both  formulae  give  identical 
results. 

We  shall  assume  now  that  the  three  cores  of  the  cable  are  con- 
nected to  a  three-phase  star-connected  generator  supplying  an 
e.m.f.  of  pure  sine  form,  the  neutral  point  and  the  lead  sheath 
being  grounded.  In  this  case  the  effective  capacity  of  each  con- 
ductor is 

C.  =  Ci-  Ci2.  (74) 

The  charging  currents  on  the  three  conductors  are  equal  to  each 
other  in  amplitude  and  differ,  of  course,  in  phase  by  120  degrees; 
that  is, 

/i  =  wCeEm  cos  coZ, 

72  =  wCeEm  cos  U*  -  -j,  . 

73    =  uCeEm  COS  f  Cot  —  ~ 

If  we  ground   one  conductor  and  connect  the  other  two  con- 


120  FORMULAE  AND  TABLES  FOR  THE 

ductors  to  a  single  phase  generator,  the  effective  capacity  of  each 
conductor  is 

Ce=Cl~C*.  (76) 


The  charging  currents  on  the  two  conductors  are  equal  and  oppo- 
site in  sign; 

/I  =    —  72   =  CeO)Em  COS  w£. 

A  three-core  cable  is  supplied  with  a  single  phase  alternating  cur- 
rent, two  conductors  connected  in  parallel  for  the  transmission  of 
the  outgoing  current  and  the  third  conductor  being  the  return,  the 
lead  sheath  grounded.  Such  an  arrangement  offers  the  somewhat 
peculiar  condition  that  for  the  same  e.m.f.  the  charging  currents 
on  the  conductors  differ  when  the  neutral  point  of  the  winding 
is  grounded  or  insulated.  That  is  to  say,  the  effective  capacities 
of  the  conductor  depend  on  the  electrical  condition  of  the  stator 
winding,  grounded  or  insulated. 

In  the  first  case  when  the  neutral  point  of  stator  winding  is 
grounded  the  effective  capacities  of  the  three  conductors,  1,  2  and 
3,  are  |  Ci,  i  Ci,  J  (Ci  —  2  C^),  respectively. 

The  charging  currents  are 

cos  orf, 


/S  =  -  i  (Ci  -  2  Cu)  wEm  COS  CO*. 

The  charging  current  on  the  inner  surface  of  the  sheath  is 

Is  =  -  i  (Ci  +  2  Ci2)  wEm  cos  co£, 
and 

/I  +  /2  +  /8  +  /.   =  0. 

It  is  evident  from  (77)  that  the  currents  in  the  outgoing  and 
return  conductors  are  not  equal. 

When  the  stator  winding  is  insulated,  the  effective  capacities 
of  the  three  conductors,  1,  2  and  3,  are 

i  (C[  -  C12),  J  (Ci  -  CB)  and  f  (C12  -  d),  respectively. 
The  charging  currents  are 

II  =  /2   =  |  (Ci  -  Ci2)  uEm  COS  0)t,\ 

/8  =  i(Cu-Ci)«tfwcos«*.         ) 
Evidently,  as  required, 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    121 

The  above  example  shows  that  the  effective  capacity  depends 
not  only  on  the  arrangement  of  the  conductors  but  also  on  the 
electrical  condition  of  the  stator  winding  with  respect  to  the 
cable,  that  is  grounded  or  insulated. 

A  list  of  the  capacities  that  can  be  obtained  from  a  three-core 
cable  is  given  in  Russell's  "  Alternating  Current  Theory  "as  follows: 

(1)  Capacity  between  1  and  2  (3  grounded)  =  J  (Ci  —  £12). 

(2)  Capacity  between  1  and  2,  3  =  f  (Ci  -  Cy. 

(3)  Capacity  between  1  and  S  (2  and  3  insulated) 

=  (Ci  -C12)  (Ci  +  2C12) 
Ci  +  C12 

(4)  Capacity  between  1  and  S,  2  (3  insulated) 

_  (Ci  -  CM)  (Ci  +  Cn) 


(5)  Capacity  between  1  and  S,  2,  3  =  Ci. 

(6)  Capacity  between  S  and  1,  2  (3  insulated) 

_2(C1-C12)(C1+2C12) 
Ci 

(7)  Capacity  between  1,  S  and  2,  3  =  2  (d  +  C12). 

(8)  Capacity  between  S  and  1,  2,  3  =  3  (Ci  +  2  Cu). 
Example.  —  Taking  the  values  of  Ci  and  C&  as  given  on  page  119, 

Ci  =  0.074  mf.  per  mile, 
Ci2  =  -  0.0119  mf.  per  mile, 

we  get 

(1)  Capacity  between  1  and  2  =  0.043  mf.  per  mile. 

(2)  Capacity  between  1  and  2,  3  =  0.0573  mf.  per  mile. 

(3)  Capacity  between  1  and  S  (2  and  3  insulated)  =  0.0695  mf. 
per  mile. 

(4)  Capacity  between  1  and  S,  2  (3  insulated)  =  0.072  mf.  per 
mile. 

(5)  Capacity  between  1  and  S,  2,  3  =  0.074  mf.  per  mile. 

(6)  Capacity  between  S  and  1,  2  (3  insulated)  =  0.1165  mf.  per 
mile. 

(7)  Capacity  between  1,  S  and  2,  3  =  0.1242  mf.  per  mile. 

(8)  Capacity  between  S  and  1,  2,  3  =  0.1863  mf.  per  mile. 


122 


FORMULAE  AND  TABLES  FOR  THE 


TABLE  XI 

VALUES  OF  C  AND  C" 

For  use  in  Formulae  (25)  and  (26) 


_d 

C 

a 

C' 

a 

i 

V  =  a' 

C 
a 

C' 
a 

2.061 

4.6647 

0.6931 

2.6 

0.8552 

0.7340 

2.0S1 

4.0891 

0.6931 

2.7 

0.8275 

0.7401 

2.0»1 

3.5134 

0.6931 

2.8 

0.8044 

0.7460 

2.031 

2.9378 

0.6931 

2.9 

0.7847 

0.7517 

2  .Oil 

2.3626 

0.6932 

3.0 

0.7677 

0.7572 

2.01 

1.7896 

0.6942 

3.1 

0.7528 

0.7625 

2.02 

1.6191 

0.6946 

3.2 

0.7396 

0.7677 

2.03 

1.5202 

0.6954 

3.3 

0.7278 

0.7726 

2.04 

.4506 

0.6961 

3.4 

0.7172 

0.7774 

2.05 

.3970 

0.6968 

3.5 

0.7076 

0.7820 

2.06 

.3536 

0.6975 

3.6 

0.6989 

0.7864 

2.07 

.3171 

0.6983 

3.7 

0.6909 

0.7907 

2.08 

.2857 

0.6990 

3.8 

0.6836 

0.7948 

2.09 

.2582 

0.6997 

3.9 

0.6768 

0.7988 

2.10 

.2337 

0.7004 

4.0 

0.6705 

0.8026 

2.15 

1.1413 

0.7040 

5.0 

0.6263 

0.8345 

2.20 

1.0775 

0.7075 

6.0 

0.6006 

0.8577 

2.25 

1.0291 

0.7117 

7.0 

0.5836 

0.8753 

2.30 

0.9911 

0.7144 

8.0 

0.5712 

0.8891 

2.35 

0.9594 

0.7178 

9.0 

0.5626 

0.9001 

2.40 

0.9326 

0.7211 

10.0 

0.5556 

0.9092 

2.45 

0.9095 

0.7245  - 

100 

0.5051 

0.9901 

2.50 

0.8892 

0.7277 

1000 

0.5005 

0.9990 

TABLE  XII 


d 

2~a' 

-* 

cosh.  A. 

d 

2a' 

*.'-. 

cosh-'^. 

1.01 

0.7031 

0.1413 

3.5 

1.9459 

1.9248 

1.05 

0.7419 

0.3149 

4.0 

2.0794 

2.0634 

1.1 

0.7885 

0.4435 

4.5 

2.1972 

2.1846 

1.2 

0.8755 

0.6224 

5.0 

2.3025 

2.2924 

1.3 

0.9555 

0.7564 

5.5 

2.3979 

2.3896 

1.4 

.0296 

0.8670 

6.0 

2.4849 

2.4779 

1.5 

.0986 

0.9622 

7.0 

2.6390 

2.6339 

1.6 

.1631 

.0470 

8.0 

2.7726 

2.7687 

1.7 

.2238 

.1232 

9.0 

2.8903 

2.8873 

1.8 

.2809 

.1929 

10.0 

2.9956 

2.9932 

1.9 

.3350 

.2569 

12 

3.1780 

3.1763 

2.0 

.3863 

.3170 

14 

3.3322 

3.3309 

2.2 

.4816 

.4255 

16 

3.4657 

3.4648 

2.4 

1.5686 

.5216 

18 

3.5835 

3.5827 

2.6 

1.6487 

1.6096 

20 

3.6888 

3.6882 

2.8 

1.7228 

1.6886 

25 

3.9119 

3.9116 

3.0 

1.7918 

1.7627 

CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS  123 


TABLE  XIII 
SPECIFIC  INDUCTIVE  CAPACITIES  OF  SOLIDS  (AIR  =  UNITY) 


Substance. 


Specific  induc- 
tive capacity. 


Authority. 


Calcspar  parallel  to  axis 7.5 

Calcspar  perpendicular  to  axis  ....          7.7 

Caoutchouc 2.12-2.34 

Caoutchouc,  vulcanized 2.69-2.94 

Celluvert,  hard  gray 1 . 19 

Celluvert,  hard  red 1 .44 

Celluvert,  hard  black 1 .89 

Celluvert,  soft  red 2. 66 

Ebonite 2.08 

Ebonite 3.15-3.48 

Ebonite 2.21-2.76 

Ebonite 2.72 

Ebonite 2.56 

Ebonite 2.86 

Ebonite 1.9 

Fluor-spar 6.7 

Fluor-spar 6.8 

Glass,*  density  2.5  to  4.5 5-10 

Double  extra  dense  flint,  density  4.5         9 . 90 

Dense  flint,  density  3.66 7.38 

Light  flint,  density  3.20 6.70 

Very  light  flint,  density  2.87 6.61 

Hard  carbon,  density  2. 485 6.96 

Plate 8.45 

Mirror 5.8-6.34 

Mirror 6.46-7.57 

Mirror 6.88 

Mirror..,  6.44-7.46 


Romich  and  Nowak. 

Romich  and  Nowak. 

Schiller. 

Schiller. 

Elsas. 

Elsas. 

Elsas. 

Elsas. 

Rossetti. 

Boltzmann. 

Schiller. 

Winkelmann. 

Wiillner. 

Elsas. 

Thomson  (from  Hertz's 

vibrations). 
Romich  and  Nowak. 
Curie. 
Various. 
Hopkinson. 
Hopkinson. 
Hopkinson. 
Hopkinson. 
Hopkinson. 
Hopkinson. 
Schiller. 
Winkelmann. 
Doule. 
Elsas. 


*  The  values  here  quoted  apply  when  the  duration  of  charge  lies  between  0.25  and  0.00005  of  a 
second.    J.  J.  Thomson  has  obtained  the  value  2.7  when  the  duration  of  the  charge  is  about  • 
of  a  second;  and  this  ia  confirmed  by  Bloudlot,  who  obtained  for  a  similar  duration  2.8. 


124 


FORMULAE  AND  TABLES  FOR  THE 


TABLE  XIII   (Continued) 
SPECIFIC  INDUCTIVE  CAPACITIES  OF  SOLIDS  (AIR  =  UNITY) 


Substance. 

Specific  induc- 
tive capacities. 

Authority. 

Guttapercha  

3.3-4.9 

Submarine  cable  data. 

Gypsum   . 

6  33 

Curie. 

Mica  

6  64 

Klemencic. 

Mica  .    . 

8  00 

Curie. 

Mica  

7.98 

Bouty. 

Mica 

5  66-597 

Elsas 

Mica 

4  6 

Romich  and  Nowak 

Paraffin 

2  32 

Boltzmann. 

Paraffin  

1.98 

Gibson  and  Barclay. 

Paraffin  

2  29 

Hopkinson. 

Paraffin,  quickly  cooled,  translu- 
cent   

1.68-1.92 

Schiller.* 

Paraffin,  slowly  cooled,  white  
Paraffin  fluid,  pasty.. 

1.85-2.47 
1  98-2  08 

Schiller. 
Arons  and  Rubens. 

Paraffin,  solid... 

1.95 

Arons  and  Rubens. 

Porcelain  

4.38 

Curie. 

Quartz  along  the  optic  axis  
Quartz,  transverse 

4.55 
4  49 

Curie. 
Curie. 

Resin 

2  48-2.57 

Boltzmann. 

Rock  salt  

18 

Hopkinson. 

Rock  salt  

5.85 

Curie. 

Selenium 

10  2 

Romich  and  Nowak. 

Shellac  

3.10 

Winkelmann. 

Shellac 

3  67 

Doule. 

Shellac  . 

2.95-3.73 

Wiillner. 

Spermaceti  

2.18 

Rosetti. 

Spermaceti 

2  25 

Felici. 

Sulphur 

3  84-3  90 

Boltzmann. 

Sulphur. 

2  24 

J.  J.  Thomson. 

Sulphur  

2.94 

Bloudlot. 

*  The  lower  values  were  obtained  by   electric  oscillations  of  duration  of  charge  about  0.0006 
second.    The  larger  values  were  obtained  when  duration  of  charge  was  about  0.02  second. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    125 


TABLE  XIV 
SPECIFIC  INDUCTIVE  CAPACITIES  OF  LIQUIDS 


Substance. 

Specific  induc- 
tive capacity. 

Authority. 

Alcohols: 
Amyl  

15-15.9 

Cohen  and  Arons. 

Ethyl  

24-27 

Various. 

Methyl  

32.65 

Tereschin. 

Propyl  

22.8 

Tereschin. 

Anilin  

7.5 

Tereschin. 

Benzene  

1.93-2.45 

Various. 

Hexane,  between  11°  and  13°  C..  .  . 
Octane,  between  13.5°  and  14°C..  . 
Decane,  between  13.5°  and  14.2°  C. 
Amylene,  between  15°  and  16.  2°  C. 
Octylene,betweenll.5°andl3.6°C. 
Decylene,  between  16.7°  C  
Oils: 
Arachid  

1.859 
1.934 
1.966 
2.201 
2.175 
2.236 

3.17 

Landolt  and  Jahn. 
Landolt  and  Jahn. 
Landolt  and  Jahn. 
Landolt  and  Jahn. 
Landolt  and  Jahn. 
Landolt  and  Jahn. 

Hopkinson. 

Castor 

4  6-4  8 

Various 

Colza 

3.07-3  14 

Hopkinson 

Lemon 

2  25 

Tomaszewski 

Neatsf  oot  

3.07 

Hopkinson. 

Olive  

3.08-3.16 

Arons  and  Rubens  ]  Hop- 

Petroleum    
Petroleum  ether  
Rape  seed 

2.02-2.19 
1.92 
2  2-3  0 

kinson. 
Various. 
Hopkinson. 
Various 

Seasame 

3  17 

Hopkinson 

Sperm 

3  02-3  09 

Hopkinson 

Turpentine  

2.15-2.28 

Various. 

Vaseline  

2.17 

Fuchs. 

Ozokerite  

2.13 

Hopkinson. 

Toluene  

2.2-2.4 

Various. 

Xylene 

2  3-2  6 

Various 

126  FORMULAE  AND  TABLES  FOR  THE 


CHAPTER  IV 
ALTERNATING    CURRENT   CIRCUITS 

IN  alternating  current  circuits  the  distribution  of  the  currents 
and  potentials  in  any  arrangement  of  circuits  is  governed  by  the 
frequency  of  the  impressed  electromotive  force,  the  resistances, 
self  and  mutual  inductances  and  capacities  of  the  various  branches 
of  the  circuit.  A  variation  in  any  of  the  electrical  constants  may 
result  in  a  change  in  the  currents,  in  magnitude  as  well  as  in 
phase,  in  all  the  branches  of  the  circuit.  Every  alternating 
current  problem  must  therefore  be  considered  separately  and 
careful  account  must  be  taken  of  all  the  electrical  constants  of 
the  circuit.  Since  there  is  a  wide  range  of  possible  arrangements 
of  circuits,  we  shall  require  a  considerable  number  of  formulae  to 
cover  all  the  more  important  cases  which  may  arise  in  practice. 

It  is  also  to  be  noted  that  the  permanent  value  of  the  current, 
which  in  the  case  of  alternating  currents  is  a  constant  periodic 
function  of  the  time,  is  reached  only  after  an  appreciable  time 
interval  when  the  circuit  is  closed.  At  the  moment  of  closing  or 
opening,  or  on  any  change  in  the  electrical  condition  of  circuits, 
a  transient  phenomena  occurs  which  is  usually  active  only  for  a 
very  short  interval.  The  formulae  for  transient  phenomena  will 
be  considered  in  Chapter  V,  while  in  this  chapter  we  shall  give 
a  collection  of  formulae  for  the  permanent  values  of  current  and 
potential  distribution  in  various  arrangements  of  circuits. 

In  this  and  the  succeeding  chapters  the  following  notation  will 
be  used : 

Notation.  — 

r  =  resistance. 
L  =  inductance. 
C  =  capacity. 
N  =  frequency, 
co  =  2  vN. 
xm  =  wL  =  inductive  reactance. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    127 

xc  =  -7  ^  =  capacity  reactance. 
coC 

x  =  xm  —  xc  reactance  of  circuit  having  inductance  and  ca- 

pacity. 

Z  =  r  +  j%  —  impedance. 
z  =  Vr2  +  x2  =  absolute  value  of  impedance. 

Y  =  -^  =  admittance. 

Li 

y  =  Vg2  -\-  b2  =  absolute  value  of  admittance. 
g  =  conductance. 
b  =  susceptance. 
„  1  r—jx 

-' 


g  =  -g—  r-  —  2  =  -g  =  effective  conductance  in  mhos,  energy 

7*     ~T~  X  £ 

component  of  current. 

/v»  /v» 

6  =  -5—  ;  —5  =  -s  =  effective  susceptance  in  mhos,  wattless 
r2  +  a;2      z2 

component  of  current. 

r  =    2  *.   ,  2  =  —  2  =  effective   resistance   in   ohms,    energy 
component  of  e.m.f  . 

s  =    o   ,   79  =  —  9  =   effective  reactance  in  ohms,  wattless 
g2  +  b2      y2 

component  of  e.m.f. 

Any  alternating  current  wave  can  be  resolved  by  the  aid  of 
Fourier's  theorem  into  a  series  of  waves,  each  one  being  a  sine 
function  in  its  variation  with  respect  to  time;  hence  in  the  dis- 
cussion of  alternating  current  problems  it  is  nearly  always  justi- 
fiable to  assume  that  the  impressed  electromotive  force  is  of 
sine  form.  If  it  is  a  complex  wave  it  can  be  resolved  into  its 
several  harmonic  components,  and  each  one  treated  as  a  separate 
source  of  e.m.f. 

If  E  is  a  periodic  function  with  respect  to  time  we  can  put  it  in 
the  following  form  : 


^3)H  ----   (1) 

The  alternating  electromotive  force  and  current  curves  which 
occur  in  practice  have  only  odd  harmonics.     A  complete  dis- 


128 


FORMULAE  AND  TABLES  FOR  THE 


cussion  of  the  theory  of  harmonic  analysis  is  beyond  the  scope  of 
this  book,  and  to  give  only  an  outline  of  the  subject  would  not 
be  of  any  practical  value.  The  reader  will  find  a  good  discussion 
on  the  general  subject  of  harmonic  analysis  in  Byerly's  Fourier's 
Theory  and  Spherical  Harmonics,  and  its  application  to  the 
analysis  of  alternating  current  curves  is  quite  fully  discussed  in  the 
"  Theorie  der  Wechselstrome,"  by  J.  L.  LaCour  and  0.  S.  Bragstad. 
In  alternating  currents  we  have  to  distinguish  between  the 
maximum  value,  the  effective  value,  which  is  the  square  root  of  the 
mean  square,  and  the  mean  or  average  value.  In  the  case  of  a 

A 

simple  harmonic  function  the  effective  value  is  —7=  where  A  is  the 

V2 

maximum  value,  and  the  mean  or  average  for  a  half  period  is  — - 

7T 

The  arithmetic  mean  for  a  whole  period  is  zero.  In  Table  XV 
we  give  the  relation  between  effective,  mean  and  maximum 
values  for  various  shapes  of  curves. 

TABLE  XV 

CHARACTERISTIC  FEATURES  OF  DIFFERENT  FORMS  OF  ALTERNATING 
CURRENT  AND  PRESSURE  CURVES 


Area  of 

Ratio  of 

Ratio  of 

Ratio  of 

Ratio  of 

squared 

Name  of  curve. 

average  to 
maximum 

effective  to 
maximum 

maximum 
to  effective 

effective  to 
average 

responding 

value. 

value. 

value. 

value. 

maximum 

value. 

Sinusoid 

0.637 

0  707 

1.414 

1.112 

0.500 

Semicircle  

0.785 

0.835 

1.198 

1.063 

0.697 

Parabolic  curve  

0.666 

0.730 

1.369 

1.096 

0.533 

Triangle  

0.500 

0.577 

1.732 

1.155 

0.333 

Approximate  rectangle. 

0.856 

0.889 

1.124 

1.038 

0.791 

Rectangle  

1.000 

1.000 

1.000 

1.000 

1.000 

In  the  case  of  complex  waves  consisting  of  several  harmonies 
the  effective  value  is** 

Eea  =  V±(#i2+#32  +  #52  +  .  -  •  ),  (2) 

where  E\,  Ez,  E$,  etc.,  are  the  maximum  values  of  the  several 
harmonic  components. 

*  See  "Theorie  der  Wechselstrome,"  von  J.  L.  LaCour  und  O.  S.  Bragstad, 
Ch.  XII. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    129 

Form  factor.  —  The  ratio  of  the  effective  value  of  a  periodic 
curve  to  its  mean  value  is  called  the  form  factor,  because  it  varies 
with  the  form  of  the  curve,  the  more  peaked  the  curve  the  greater 
the  value  of  the  form  factor.  For  a  pressure  curve  the  form 
factor  is  given  by  the  expression 


(3) 
Edt 


For  a  sine  curve  the  form  factor  is 
1        2          TT 


V2  '  *      2\/2 


=  1.112. 


Curve  factor.  —  The  ratio  of  the  effective  value  to  the  maxi- 
mum value  of  the  fundamental  harmonic  is  called  the  curve 
factor. 

E  m 


As  an  illustration  consider  the  case  of  an  e.m.f.  curve  having  three 
upper  odd  harmonics.  The  maximum  values  of  the  fundamental 
and  the  harmonics  are 

E!  =  100,     #3  =  40,     E,  =  20,     E7  =  10. 
The  effective  value  is 


#efl  =  V  (100)2  +  (40)2  +  (20)2  +  (10)2  =  110. 
The  curve  factor  is 


e_ 

IT    lop 

Alternating  Current  Circuits.  —  When  an  electromotive  force 
of  sinusoidal  form,  E  cos  ut,  is  impressed  on  a  circuit  containing 
inductance  and  resistance,  it  generates  a  current  in  the  circuit. 


C°S     "*  - 


the  current  lagging  behind  the  electromotive  force  by  the  angle 


130 


FORMULAE  AND  TABLES  FOR  THE 


Example.  — 

r  =  5ohms,      L  =  0.01  henry,     w  =  2^X60,    E  =  500  volts. 
xm  =  coL  =  3.73. 
500 


V(3.77)2  +  (5)2 


=  80  amp. 


tan 


^  =  0.754,     0  =  37°  3'. 


The   maximum   value  of  the  current  is  80  amperes  and  it  lags 
37°  3'  behind  the  e.m.f. 

If  the  circuit  contains  capacity  ^reactance  and  resistance,  the 
current  is 

Tjl 

cos  (ut  +  ^), 

(6) 


the  current  leading  the  electromotive  force  by  the  angle  \f/.  The 
voltage  on  the  condenser  is 

xcE 


Example.  — 

r  =  5  ohms,     C  =  0.002  farads,    co  =  2  IT  X  60,     E  =  500  volts. 

xc  =  4  =  1-324, 


E 


500 


V(1.324)2 


=  ±^P  =  0.265,     I  =  14' 
o 


=  96.7  amp., 


51'. 


The  current  leads  the  e.m.f.  by  the  angle  14°  51r.     The  voltage 
across  the  condenser  is 

1.324  X  500  . 

e  —     ,  =  128  volts. 

V  (1.324)2  +  (5)2 

When  the  circuit  contains  both  capacity  reactance  and  inductive 
reactance. we  have 

E 

*  vA^1-* 


//  x 

V  (xm  -  xc) 


tan  7  = 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    131 
When  xm  =  xc,  equation  (8)  reduces  to 


7 

I  =  —  cos  wt. 


(9) 


The  circuit  acts  as  if  there  is  no  reactance,  and  it  is  said  to  be  in 
resonance  with  the  frequency  of  the  impressed  electromotive 
force.  The  condition  for  resonance  then  is 


or 


or 


J^ 
«C' 
1 


1, 


VLC 


(10) 


30 

28 
26 
24 
22 
20 
18 

4-> 

I" 
12 

10 


$ 


/ 


Curve     I,  R  =  10  Ohms 

.     «•         II,  R=  5      •• 

•«       HI,  JJ=3.5     - 

IV,  #=2.0     « 


n 


ss 


\\ 


0   2   4   6   8  10  12  14  16  18  20  22  24  26  28  30  32  34  86  38  40 

i 

FIG.  36.  —  Resonance  curves  for  different  values  of  resistance  in  the  circuit. 

In  any  circuit  containing  inductance  and  capacity  it  is  always 
possible  to  adjust  those  constants  so  as  to  produce  resonance 
condition  for  a  given  frequency.  For  a  given  electromotive  force, 
the  current  rises  as  the  resonance  condition  is  approached,  which 
is  more  pronounced  the  smaller  the  resistance  in  the  circuit. 
In  Fig.  36  are  plotted  a  series  of  resonance  curves  for  different 
resistances. 


132  FORMULAE  AND  TABLES  FOR  THE 

The  voltage  across  the  condenser  is 

xcEsm(at-y) 
~ 


For  resonance  condition  and  small  resistance,  the  voltage  across 
the  condenser  may  exceed  many  times  the  voltage  of  the  generator. 
Example.  — 

w  =2irX60    C  =  20mf. 


1325 

=  66.2  #  sin  o>*. 


The  voltage  across  condenser  is  about  6600  volts,  66  times  the 
electromotive  force  of  the  generator. 

If  the  electromotive  force  is  a  complex  wave,  that  is,  of  the  form 


E  =  EI  cos  (co£  +  «i)  +#3  cos  (3  ut  +  as)  +  E$  cos  (5  coi  +  #5) 
the  current  generated  in  the  circuit  is 

,  __  EI  cos  (co<  +  cti  —  7i)    ,  E8  cos  (3  at  +  «3  —  73) 
V(zw-zc)2  +  r2 


^5  COS  (5  (0^  +  <^5~  7s)     i 


« ^m      "o"  5  xm      ^~ 

-^ -»  tan 73  = »    tan 75  = •••    (12) 


It  may  happen  that  the  inductance  and  capacity  of  the  circuit 
are  of  such  magnitudes  as  to  bring  the  circuit  into  resonance  for 

/7* 

some  harmonic  of  the  wave,  for  instance  we  may  have  3  xm  —  -£ 

o 

=  0,  that  is,  the  circuit  is  in  resonance  for  the  third  harmonic.  In 
that  case  the  current  produced  by  the  third  harmonic  in  the  elec- 
tromotive force  wave  may  be  larger  than  that  produced  by  the 
fundamental,  though  the  amplitude  of  the  e.m.f .  in  the  fundamental 
may  be  much  larger  than  that  of  the  third  harmonic. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    133 

Example.  — 

C  =  7.77  mf. 
L  =  0.1  henry. 
co  =  2  TT  X  60. 
Ei  =  100,     Ez  =  10,     r  =  10  ohms. 


=  0.33  amp. 


10 
=       =  lamp. 


The  current  due  to  the  first  harmonic  of  the  e.m.f.  curve  is  three 
times  as  large  as  the  current  due  to  the  fundamental,  though  the 
maximum  value  of  the  fundamental  in  the  e.m.f.  curve  is  ten  times 
as  large  as  the  first  harmonic. 

If  there  are  several  impedances  Zi,  22,  23  •  •  •  connected  in  series 
in  the  circuit,  the  resultant  is  the  algebraic  sum  of  the  separate 
impedances.  Denoting  by  z~t  the  total  impedance,  we  have 

2*  =  '3i  +  3z  +  23  +  •  -  •  =  (TI  +  r2  +  r3  +  •  •  •  ) 

i  +  x2  +  x3+  .  .  .  ),  (13) 


and  the  absolute  value  of  the  current, 

/  =  -  JL  •        (14) 

•  •  •  )2  +  (ri  +  r2  +  r3  +  -  -  -  )2 


The  e.m.f.  En  across  any  impedance  coil,  say  the  nth  coil,  is 

-v/r  2  4-  r  2 
En  =  ^     ^^  '  (15) 

x2  +  z3  +  •  •  •  )2  +  (ri  +  r2  +  r3  +  •  •  •  )2 


Parallel  Circuits.*  —  Two  circuits  in  parallel,  each  having 
resistance  and  inductance.  Denoting  the  reactances  and  im- 
pedances of  the  two  branches  by  Xi,  x2  and  2i,  Z2  respectively,  the 

*  Alex.   Russell,    "Theory  of   Alternating  Currents,"    Vol.   I,   Ch.  VII. 
Andrew  Gray,  "Absolute  Measurements  in  Electricity  and  Magnetism," 
Vol.  II,  Pt.  I,  Ch.  IV. 


134  FORMULAE  AND  TABLES  FOR  THE 

currents  in  the  two  branched  circuits  and  the  main  circuit  are 

E 

1  1  —  —cos  (co£  —  <£i), 

2i 


22 

E 

/O  =  -  —  COS  ((jit  -  00), 


=          tan  02  =  ->     tan00 
n  r2 

or  we  may  write 


2 
21      22 


E  cos  M  -  00)      £7        '..  . 

o  =  —  /  o  .         =  rcos  W  ~  w» 

V  r02  +  a;02         2« 
where 

jTi   i  J5 

2  2     '02 

r0  =  r  -  rr  —  ;  -  r-ris  the  equivalent  resistance  of  the 


circuit  and 

2  +    2    is  the  ecluivalent  react- 


ance  of  the  circuit.     If  the  resistances  and  reactances  of  the  two 
branched  circuits  are  equal,  the  above  values  reduce  to 

r0  =  \  r,        x0  =  i  x. 
Example.  — 

Xi  =  5  ohms,  ri  =  3  ohms,  x2  =  3  ohms,  r2  =  5  ohms. 

2l  =  z2  =  5.83,        E  =  500  volts. 

E 
Ii  =  —  =  85.8  amp., 

2l 

72  =  —  =  85.8  amp., 

22 


=  tan-1      =  59°  3', 


_^  __  3>~~  — 
02  =  tan-i      =  31°, 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    135 


XQ  = 


5  +  3 
34 


17 


17 


+  (A)2       8 

500 

=  166.4  amp., 


=       -1-  = 


=  tan 


=  tan-1  1  =  45°. 


The  current  in  the  main  circuit  is  less  than  the  sum  of  the  currents 
in  the  two  branches,  and  this  is  because  the  phase  angles  in  the 
two  branches  are  different. 


-AAAAAT 


FIG.  37. 

Two  Branch  Circuits,  One  having  Resistance  Only  and  the 
Other  Inductance  Only.  —  One  branched  circuit  has  a  resistance 
r2  and  the  other  an  inductance  LI  of  negligible  resistance,  the 
main  circuit  inductance  and  resistance  being  r0  and  L0  (see  Fig.  37). 
The  total  impedance  of  circuit  A  —  B  is  given  by  the  following 
expression : 


This  is  a  maximum  when 
co2Li  (L2  +  2  LC 


2  rc 


.     (18) 


For  the  value  of  r2  given  by  equation  (18),  the  current  in  the  main 
circuit  is  a  minimum.  Hence  shunting  an  inductance  coil  with 
a  resistance  sometimes  increases  the  apparent  resistance  of  the 
circuit,  a  result  which  has  been  noticed  in  practical  work. 


136 


FORMULAE  AND  TABLES  FOR  THE 


Example.  — 

Lo  =  L2  =0.1  henry,'  r0  =  50  ohms,   r2  =  129  ohms,    co  =  2w  X  60. 

By  (17)  we  have 

Z2=  (68.1)2  +  (61.75)2  =  8450.67, 
Z  =  91.9  ohms. 

If  the  inductance  L%  were  not  shunted  by  any  resistance,  the 
impedance  of  the  circuit  would  be 

Z2  =  co2  (Li  +  L0)2  +  r02  =  (75.4)2  +  (50)2  =  8185.16, 
Z  =  90.5  ohms. 

Shunting  the  inductance  by  a  resistance  increases  the  impedance 
of  the  circuit  instead  of  decreasing  as  we  would  ordinarily  expect. 
A  System  of  Branched  Circuits  in  Parallel  Each  having  Resis- 
tance and  Inductance.  —  Let  z\,  %,  0,3 ...  denote  the  impedance 
of  the  respective  branches  and  ZQ  the  total  impedance  of  the  cir- 
cuit. The  currents  in  the  respective  branches  and  the  main  circuit 
are  as  follows: 

TjJ 

/!  =  --cos(coZ  -  <£i), 


1 2  =  —COS  (ut 
Z2 


E 

In=  —  COS  (otf 

zn 

-pi 

IQ  =—  COS  (at 


where 


=  W2 


(19) 


X0 


V  +  6 


=  equivalent  resistance, 


=  equivalent  reactance, 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    137 
and 


k=n 


n  =  the  number  of  branched  circuits  in  parallel. 

If    the    resistances    and    inductances    respectively  of    all  the 
branched  circuits  are  equal, 


and 


1    ,    1       n2(r2  +  x2) 


and 


E 


v/i 


cos  (orf  —  00), 


(r2  +  x 


(20) 


that  is,  the  main  current  is  n  times  the  current  in  any  of  the 
branches. 

Jo  •     /i 


E 


FIG.  38. 

Inductance  and  Resistance  Shunted  by  Condenser.*  — 
Let  /i  denote  the  current  in  inductance  branch,  72  the  cur- 
rent in  capacity  branch  and  70  the  main  current  and  E  cos  ait  the 
impressed  electromotive  force.  The  currents  in  the  two  branches 

*  See  "Theory  and  Calculations  of  Alternating  Current  Phenomena,"  3d 
edition,  Ch.  VIII,  by  C.  P,  Steinmetz. 


138      *          FORMULAE  AND  TABLES  FOR  THE 

and  the  main  circuit  are 
E  cos  (ut  —  <j>) 


I2  =  Exc  cos  co£, 


(21) 


_ 

o/c       o/m    T  i         «**  /oo^^ 

tan  ^  =  -  =  --          (22) 


r2      I     r     2 
I        ^^  U/jji 

When  -  =      2*_7    2  (23) 

*^c          ^m       I     ' 

the  circuit  is  in  resonance,  that  is,  the  main  current  is  in  phase 
with  the  electromotive  force  and 

Er 

lo  =  -^2 — 2  cos  ut.  (24) 

T      \    Xm 

From  (23)  it  is  obvious  that  for  this  arrangement  the  resonance 
condition  depends  not  only  on  the  inductance  and  capacity, 
but  also  on  the  resistance.  For  every  increase  in  r,  xc  must  be 
increased  and  the  capacity  decreased  to  maintain  the  resonance 
condition. 

By  varying  xc  we  can  maintain  the  current  in  the  main  circuit 
in  phase  with  the  e.m.f.  for  any  variation  in  the  resistance  of  the 
receiving  circuit.  As  an  illustration  we  assume  a  value  of  xm  =  15, 
and  calculate  the  values  of  xc  required  to  maintain  the  resonance 
condition  in  the  circuit  for  values  of  r  from  1  to  10  ohms.  The 
results  are  plotted  in  the  curve  given  in  Fig.  39. 

The  e.m.f.  EQ  across  the  resistance  r  is 

Er 

.cos(co£-0).  (25) 


,'  +  r' 

It  is  also  to  be  noted  that  in  such  an  arrangement  of  circuits,  at 
resonance  condition,  the  currents  in  the  branches  are  larger  than 
the  currents  in  main  circuit. 
Example.  — 

xm  =  25,     r-10,     f  -  TowSlTnyS  =  °-0345'    #  =  100  volts, 

xc      \  •"<->/   -f-  v^-w 

100  3.7  amp., 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    139 

72  =  100  X  0.0345  =  3.45  amp., 
100  X  10 


625OO 

The  currents  in  the  branch  circuits  are  considerably  larger  than 
the  current  in  the  main  circuit. 


21 

20 

19 

-1 

0 

« 
17 

16 
15 

/ 

/ 

Curve  giving  Variation  in  Xc 
with  r  so  as  to 
Maintain  Main  Circuit 
Current  in  Phase  with  e.m.f. 

/ 

' 

^ 

/ 

^ 

/ 

x 

^ 

x 

7 

^x 

^ 

/ 

i 

x 

X 

x 

? 

_»••' 

^ 

^ 

,  • 

.  —  —• 

,~—- 

1             2             3             4              5      r     6             7              8             8            1C 
FIG.  39. 

FIG.  40. 

Inductive  Load  Shunted  by  Condenser.  —  Assume  an  elec- 
tromotive force  of  sine  form  impressed  on  the  circuit  and  a  react- 
ance x0  in  series  with  alternator  (see  Fig.  40).  Denote  by  /i  the 
current  in  resistance  branch,  72  the  current  in  condenser,  70  the 
main  current,  E  the  voltage  of  alternator  and  EQ  the  voltage  on 
the  receiving  circuit. 


140  FORMULAE  AND  TABLES  FOR  THE 

The  distribution  of  the  currents  and  voltage  are  as  follows: 
Exc 


(xe  - 


tan 


xc(x0 


(xc  - 


/2=- 


E  Vrt2  + 


V  j  zc'(z0+  Zi)  -  xixol  2  +  ri2  (xc  - 


cos  (coi  —  </>i), 


—  0i  4- i/O- 


,      xl 
tan  y  =  —  > 

T 


cos  (otf  — 


tan  7  = 


Exc    n2  + 


(26) 


If  we  make  xc  =  Xi  —  XQ  =  x  the  denominators  in  the  above 
equation  reduce  to  x  and  the  equations  simplify  to 


1 1  =  —  sinotf, 
x 


(27) 


From  these  equations  we  can  draw  the  interesting  conclusion  that 
for  a  constant  value  of  E  the  current  in  the  receiving  circuit  /i 
is  constant,  and  independent  of  any  variation  in  the  load  r\. 
In  other  words  this  arrangement  of  circuits  acts  like  a  transformer 
converting  a  constant  potential  into  a  constant  current. 

Circuits  in  Parallel.  —  Two  branch  circuits  in  parallel,  one 
having  inductance  and  resistance  and  the  other  inductance 
capacity  and  resistance  (see  Fig.  41).  Denote  the  current  in 
branch  (1)  by  7i,  that  in  branch  (2)  by  72  and  the  current  in 
main  circuit  by  /o;  then 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    141 

E 


Sl 


£2 


tanT= 


-  (28) 


FIG.  41. 
The  circuit  is  in  resonance  when 

n        I  o    T 


=  o. 


(29) 


Putting  for  brevity 


+ 


=  k,  equation  (29)  becomes 


+  fcr 


and 


-  liVT11 


2/b 


(30) 


The  resonance  condition  depends  on  the  resistance  of  the  branch 
circuits  as  well  as  their  inductances  and  capacities.  If  the  con- 
stants of  the  circuits  are  of  such  magnitudes  as  to  make  4  &2r22 
greater  than  unity,  the  value  of  x2  as  given  by  (30)  is  an  imaginary 
quantity,  that  is  to  say  it  is  not  possible  to  produce  resonance  in 
the  circuit.  If  4  &2r22  is  less  than  unity,  x%  has  two  distinct 


142 


FORMULAE  AND  TABLES  FOR  THE 


values.     The  system  is  double  periodic;   for  the  same  constants 
we  may  have  resonance  at  two  different  frequencies. 
As  an  illustration,  let  k  =  0.1,  r2  —  5. 

-  1  ±  Vl  -  0.2  OK 

**  =       2x0.1  -95; 

that  is,  if  the  frequencies  are  such  as  to  make  x%  =  —  5  or  —  95, 
the  circuit  is  in  resonance  and  the  current  in  the  main  circuit  is 


/o  = 


r22     \ 
i?+^j 


cos  co£. 


(31) 


The  Branch  Circuits  as  Well  as  the  Main  Circuit  having  In- 
ductance, Capacity  and  Resistance. —  The  currents  in  the  branch 
circuits  are  denoted  by  I\  and  72  respectively,  and  the  current  in 
the  main  circuit  by  7o- 


Vc2  +  d2 


E 


+ 


cos  (ut  —  <t>  + 


Vc2  +  d2 


cos(o>Z  —  0  + 


(32) 


XO=LO u- 


7QU 


where 

tan0=-, 


and 


c  =  r0 
d  =  n 


FIG.  42. 


— ,    tan  \f/z 
TZ 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    143 


When  the  inductances  and  capacities  are  so  adjusted  that  each 
branch  is  separately  in  resonance,  we  get 

Xi  =    XZ  =    XQ  =  0, 

c  =  TO  (ri+  r2)  +  rir2,  d  =  0, 

and 
V, 


/2   = 


Er2 

TO  (r.  +  r2)  +  ri 
En 

r<>  (n  +  r2)  +  n 
E  (n  +  r2) 

—  p.ns  t,\t. 

(33) 


TO  (n  +  r2)  + 

Mutual  Inductance  Between  the  Branched  Circuits.*  —  We 

shall  now  consider  the  case  of  two  branched  circuits,  in  which 


t 


AT 


FIG.  43. 

there  is  also  mutual  inductance  between  the  two  branches.  De- 
noting the  currents  in  the  branched  circuits  by  /i  and  72  respec- 
tively and  the  current  in  the  main  circuit  by  I0  we  have  the 
following  formulae  for  the  currents  in  the  circuits : 


Vfco2(M2-L1L2)+r1r2p+a,2(L1r2+L2r1)2 


cos  coc  — 


EV(Li- 


VJco2(M2-L1L2)+r1r2j2+co2(L1r2+L2r1>2 
E 


-  2 


r2) 


Vjco2(M2-L1L2)+7-1r2j2+co2(L1r2+L2r1)2 

CO  (LiT-2  + 


cos(coi  — 


(34) 


tan  \I/2  = 


tan 


r2. 


*  J.  J.  Thomson,  "Recent  Researches  in  Electricity  and  Magnetism,"  Ch. 
VI;  Lord  Rayleigh,  Phil.  Mag.,  May,  1886. 


144 


FORMULAE  AND  TABLES  FOR  THE 


It  is  interesting  to  note  that  in  such  an  arrangement  of  circuits, 
the  currents  in  the  branched  circuits  may,  under  some  circum- 
stances, be  considerably  larger  than  the  current  in  the  main  cir- 
cuit. Considering  only  the  amplitudes  of  the  currents,  we  have 


V(L2  -  M)2o> 


L2  -  2 


(n  +  r2)s 


h 

/o 


\/(Li  -  M)2o;2 


L2  -  2  M)2co2  +  (n  +  r2) 


(35) 


(36) 


If  we  neglect  the  resistances  of  the  two  branches  compared  with 
the  reactances  we  have 

7i=        L2-M 


and     -^  = 


(37) 


As  an  extreme  case  we  may  take  the  coefficient  of  coupling  equal 
to  unity,  that  is,  M  =  VLiL2,  and 


Li+L2-2 


-VL, 


Li  +  L2  -  2 


i  -  VL2 


(38) 

/- 


If  LI  and  L2  are  not  much  different  from  each  other  the  denomi- 
nators in  equations  (38)  are  very  small  and  consequently  the 

ratios  -^  and  ~  are  large;  that  is,  the  currents  in  the  branch  cir- 

-/o          -to 

cuits  are  considerably  larger  than  the  current  in  the  main  circuit. 
As  an  illustration  let  us  take  the  following  constants: 

Li  =  0.1  henry,     L2  =  0.01  henry,     M  =  0.03  henry. 


/o 


0.02  _  72      0.07 

0.05  '     /0      0.05 


1.4. 


The  current  in  branch  2  is  1.4  times  the  current  in  the  main  cir- 
cuit. 

Power  Factor.*  —  If  an  alternating  e.m.f., 

e  =  E  V2  cos  at, 

*  See  C.  P.  Steinmetz  "Theory  and  Calculation  of  Alternating  Current 
Phenomena,"  3d  edition,  Ch.  XXVII;  also  "Theorie  der  Wechselstrome," 
von  J.  L.  LaCour  und  O.  S.  Bragstad,  Ch.  XII. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    145 

produces  in  a  circuit  a  current, 

i  =  lV2  cos  (at  =t  0), 
where  0  is  the  phase  angle  lagging  or  leading,  the  power 

p  =  ei  =  EIlcos 0  +  cos  (2 <at  =b  0) { .  (39) 

The  average  value  of  the  power  is 

P=  EIcos<j>  =  7 


Ir 

COS0   =  -g' 

The  factor  cos  0  is  called  the  power  factor. 
Substituting  (40)  in  (39)  we  get 


40) 


p=Pl+-  -^   •  (41) 

[  COS0 

In  an  ordinary  alternating  current  circuit  the  power  fluctuates 
between  the  values 

and    PJl —I-  (42) 

[  COS0J 

When  the  phase  angle  is  zero,  cos  0  =  1,  the  fluctuation  of  the 
power  is  between  2  P  and  0.     If  the  phase  angle  is  90  degrees, 

p  =  P  (1  +  cos  2  o>Z), 
and  the  average  or  effective  power  P  =  0. 

If  the  pressure  curve  is  not  a  simple  harmonic  curve  but  con- 
tains also  harmonics,  the  average  power  supplied  to  the  circuit 
is  given  by  expression, 

P  =  EJi  COS  0i  +  #3/3  COS  03  +  EJ&OS  05  +    -    -    - 

=  £7cos0,  (43) 

where  E  and  I  are  the  effective  values  of  the  e.m.f .  and  the  current, 
and 


.,     (44) 


COS2  0!  +    •         COS2  03  +  COS2  05  + 


f) 


146  FORMULAE  AND  TABLES  FOR  THE 

where  <£i,  <£3,  <£5,  etc.,  are  the  phase  angles  corresponding  to  the 
different  harmonics. 

Power  Transmission.*  —  In  alternating  current  transmission 
problems  account  must  be  taken  of  the  line  reactance  which  is  of 
great  importance;  the  efficiency  of  transmission  and  the  voltage 
regulation  depend  on  the  electrical  constants  of  the  line.  In  the 
complete  theory  of  long-distance  transmission  we  must  allow  for 
the  effect  of  the  distributed  capacity  and  inductance  of  the  line. 
This  is  worked  out  fully  in  Chapter  VI.  In  this  section  we 
shall  assume  that  the  resistance,  inductance  and  capacity  of  the 
line  are  localized;  that  is,  we  replace  the  total  inductance  and 
capacity  of  the  line  by  an  inductance  coil  and  condenser  of  equiv- 
alent value.  This  method  gives  fairly  accurate  results  for  short 
transmission  lines,  but  for  long  lines  the  more  complete  formula 
given  in  Chapter  VI  will  have  to  be  used. 


x       r 


FIG.  44. 

As  a  first  approximation  we  shall  neglect  the  capacity  of  the  line 
entirely,  and  we  shall  also  assume  that  the  load  is  non-inductive. 
Let 

x  =  reactance  of  line, 
r  =  resistance  of  line, 
r0  =  resistance  of  receiving  circuit, 
Eg  =  e.m.f.  at  generator, 
Er  =  e.m.f.  at  receiving  circuit^ 
I  =  current. 

The  current  in  the  circuit  is  given  by  the  following  formula : 

7  =  —=Jk=      =  =  ^«  (46) 

Vx*  +  (r  +  r0)2       r0 

*  See  C.  P.  Steinmetz,  "Theory  and  Calculation  of  Alternating  Current 
Phenomena,"  Chapter  IX;  also  "Theorie  der  Wechselstrome,"  von  J.  L. 
LaCour  und  O.  S.  Bragstad,  Ch.  IV. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    147 
The  ratio  of  e.m.f.'s  at  receiving  circuit  and  generator  is 

a  =  —  =  r°  (47) 


Eg          z2  +  (r  +  r0) 
The  power  delivered  in  the  non-inductive  receiving  circuit  is 

P  =  ErI=    2_/;V_?     „•  (48) 

z2  +  (r  +  r0)2 

If  #,  a;  and  r  are  fixed,  the  power  supplied  varies  with  r0  and  it 
has  a  maximum  value  when 

ro  =  Vx2  +  r2  =  z.  (49) 

Substituting  the  value  of  r0  from  (49)  into  (48)  we  get 

Pmax  =  2(z  +  r)'  (50) 

For  maximum  power  supply  at  the  receiving  circuit  the  ratio  of 
e.m.f.  at  receiver  and  generator  is  given  by 

Er 
^max  ~~  "rT 


The  total  power  supplied  by  generator 

P«  =E0Icos<j>  =1j^>  (52) 

The  efficiency  for  maximum  power  supplied  to  receiving  circuit 

is 

Pmax=        *_  . 


For  comparison  we  will  give  here  a  set  formulae  for  a  non-induc- 
tive line  corresponding  to  equations  (46)  to  (53),  which  are  as 
follows: 


Power  supplied  to  receiving  circuit  is 

(56) 


148  FORMULAE  AND  TABLES  FOR  THE 

and  it  is  a  maximum  when  r  =  TO- 


Total  power  supplied  is 


(57) 


••-r  +  r.-27' 
The  efficiency  for  maximum  power  supply  to  receiving  circuit  is 

P  1 

•*•  max  -*• 


120 
110 
100 
90 
80 
70 
60 
50 
40 
30 
20 
10 

100 
90  • 
80  ' 
70 
60  | 
50  l 
40 
30 
20 
10 

-— 

*^- 

_—  — 

^ 

— 

X 

^ 

^ 

/ 

/ 

Sx^ 

^ 

s., 

^ 

/ 

*$> 

/ 

X 

\ 

^ 

/ 

/ 

\ 

7 

/ 

/ 

^ 

^ 

«^~ 

^~- 

—  — 

.  — 

1 

/ 

^ 

^^ 

"^ 

I 

/ 

f 

s 

^ 

1  —  *. 

•*^ 

~-^^ 

1 

/  f 

/ 

Inductive  Line,  r=4Ohms 

Curve,   I       Efficiency 
II       Current 
»«      III      Voltage  Er 
u       IV       Power 

f^ 

, 

/ 

7 

// 

iff 

f 

700 
650 

'geoo 

^550 

1  500 
§450 

2  400 
«350 
^300' 
>250 

1200 
150 
100 
50 

0123456789    10  11  12   13  14  15   16  17  18  19   20 

r 
FIG.  45.  —  Power  transmission  characteristic  curves. 

As  an  illustration  we  give  here  a  set  of  curves  showing  the 
variation  with  the  load  of  /,  Er,  TQ  and  Pr  for  an  inductive  line  of 
the  following  constants: 

E  =  1000  volts,    r  =  4  ohms,    x  =  8. 

Inductive  Load.  —  Let  Xi,  r\  represent  the  reactance  and  resist- 
ance of  transmission  line,  xz,  r2  the  reactance  and  resistance  of 
receiving  circuit,  Eg  =  e.m.f.  of  generator,  Er  =  e.m.f.  at  receiving 
circuit. 

7TT  Tjl 

T  &a  &r 


+  r2) 


(59) 


r2 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    149 
The  phase  angle  fa  between  Eg  and  I  is  given  by 

Xi  +  %2 

tan  fa  =  — -r — 
and  the  phase  angle  fa  between  Er  and  I  is  given  by 

*v2 

tan  fa  =  —  • 
We  may  put  equation  (59)  in  the  following  form : 

I  =  E  \/ ^2  +  ^2 (60) 

0  V  (1  +  rift  +  £i62)2  +  (sift  -  ribz)2 
and 

«=^=— =  *a  _  (61) 

where 

ft  =     o  ,2 — 5  =  effective  conductance  of  receiving  circuit, 

r22  +  #22 

r  =  effective  susceptance  of  receiving  circuit. 


2  _ 

The  power  supplied  to  receiving  circuit, 

P  =  Er*g2  =  Eg*g«**,  (62) 

and  it  has  its  maximum  value  when 


ft  =  Vft2  +  (bi  +  62)2;  (63) 

and  for  this  value  of  ft  we  have 

«max  =    x          1       =-  (64) 

V  2  ft  (ft^i2  +  rj 

The  maximum  power  that  can  be  supplied  to  the  receiving  cir- 
cuit is 

E  2 

Pmax  =  o  /         /    . N  •  (65) 

2  (ft2i2  +  n) 

For  constant  susceptance  62  the  efficiency  has  its  maximum  value 
when  02  =  &2  and 


(66) 

If  we  do  not  neglect  the  capacity  of  the  line,  the  problem  be- 
comes somewhat  more  complex,  but  it  may  be  simplified  very 
materially  by  the  following  considerations : 


150  FORMULAE  AND  TABLES  FOR  THE 

In  engineering  practice,  power  transmission  problems  frequently 
present  themselves  in  the  following  way.  Given  the  values  of 
the  voltage  and  current  and  their  phase  relation  at  the  receiving 
circuit,  what  will  be  the  corresponding  values  of  voltage  and 
current  at  the  generator  end,  the  power  transmitted,  efficiency, 
etc.?  And  vice  versa,  if  the  e.m.f.  and  current  are  given  at  the 
generator  end  what  will  be  the  corresponding  values  at  the  re- 
ceiving end?  In  other  words  given  the  constants  of  the  line 
and  electrical  conditions  at  one  end  of  the  line  to  determine  all 
the  other  factors  of  the  problem. 

As  stated  before,  the  exact  solution  of  the  problem  necessitates 
taking  into  account  the  effect  of  the  distributed  inductance  and 
capacity  of  the  line  which  are  worked  out  in  Chapter  VI.  For 
short  lines,  however,  we  can  get  fairly  accurate  results  by  assuming 
localized  inductance  and  capacity. 

Different  formulae  may  be  obtained  depending  on  the  assump- 
tions made  regarding  the  capacity  distribution,  and  we  may  have 
the  following  cases: 

(1)  Capacity  of  line  neglected  entirely. 

(2)  Entire  capacity  of  line  shunted  across  the  middle  of  the 
line. 

(3)  Half  of  line  capacity  shunted  across  at  each  end  of  the  line. 

(4)  Four-sixths  of  line  capacity  shunted  across  the  middle  of 
the  line  and  one-sixth  of  line  capacity  at  each  end  of  the  line. 

We  shall  give  here  formulae  corresponding  to  the  first  three 
cases,  and  work  out  some  examples  to  compare  the  closeness  of 
approximation  of  the  above  three  cases.*  The  fourth  case  was 
not  considered  for  the  reason  that  the  formulae  are  somewhat 
complex  and  the  results  obtained  by  this  method  are  not  more 
accurate  than  those  obtained  by  the  method  considered  in  case 
(3).  It  is  not  considered  advisable  to  multiply  the  number  of 
formulae  unnecessarily. 

*  Ch.  P.  Steinmetz,  "Theory  and  Calculation  of  Alternating  Current 
Phenomena,"  3d  edition,  Chap.  XIII. 

F.  A.  C.  Perrine  and  F.  G.  Baum,  The  Use  of  Aluminium  Line  Wire  and 
Some  Constants  for  Transmission  Lines,  Trans.  Am.  Ins.  E.  E.,  May,  1900. 

P.  H.  Thomas,  Output  and  Regulation  in  Long  Distance  Lines,  Trans.  Am. 
Ins.  E.  E.,  June,  1909. 

P.  H.  Thomas,  Calculation  of  High  Tension  Lines,  Trans.  Am.  Inst.  E.  E.t 
June,  1909. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    151 

We  shall  use  the  following  notation: 

Eg  =  Eg'  ±  jE0"  =  e.m.f  .  at  generator, 
I0  =  lg    ±jlg"   =  current  at  generator  end, 
Er  =  Er   ±  j#r"  =  e.m.f  .  at  receiving  end  of  line, 
Ir  =  Ir    ±  j7r"  =  current  at  receiving  end  of  line. 

The  double  sign  in  the  above  notation  is  to  indicate  whether 
the  phase  angle  is  leading  or  lagging;  the  plus  sign  meaning  phase 
angle  leading,  and  the  minus  sign  phase  angle  lagging.  Suppose 
the  voltage  and  current  are  given  at  the  receiving  end  of  the  line 
and  the  phase  angle  of  the  current  with  respect  to  the  voltage 
is  leading,  then, 

Er  =  Er', 


If  on  substituting  these  values  in  the  formulae  we  obtain  for  the 
e.m.f.  and  current  at  generator  end, 

Eg=Eg'-jEg", 


it  would  mean  that  the  e.m.f.  at  generator  end  lags  behind  the 

•pi  n 

e.m.f.  at  receiving  end  by  an  angle  tan"1-^-,  ,  while  the  current 


I  " 
at  generator  leads  the  e.m.f.  at  receiving  end  by  an  angle  tan"1  ~r  • 

LQ 

Case  I.  Line  capacity  neglected  entirely. 

Suppose  that  the  admittance  of  receiving  circuit  g  ±  jb  and  Er 
are  given,  we  have 

Ir=Er(g±jb)=Ir'±jI,". 
Eg  and  Ig  have  the  following  values: 

/.   =  Ir, 

Eg  =  \Er±  Leo//'  +  rl/l  +j \LuIr'  ±  rlr' 

Where  a  double  sign  appears  in  the  above  equation,  the  upper  sign 
is  to  be  used  when  the  current  at  the  receiver  end  is  leading  and 
the  lower  sign  when  the  receiving  current  is  lagging. 

If  the  voltage  and  current  are  given  at  the  generator  end,  that 
is  Eg  and  //  ±  jlg"  given,  we  have 

=  (GQ} 

TJ<  f  T71       i      T      T   If  T  f  t  •   c  T      T   /     i         T//I!  \"'sJ 

Er  =  \Eg±L(*I9    -rI0  I  -j 


152  FORMULAE  AND  TABLES  FOR  THE 

The  double  sign  in  the  terms  of  equation  (69)  has  the  same  sig- 
nificance as  in  (68). 

Example.  —  Power  to  be  transmitted  over  a  three-phase  line  50 
miles  long  using  hard-drawn  stranded  copper  wire  No.  000,  tri- 
angularly spaced  10  feet  spacing  between  wires.  Frequency  60 
cycles.  At  the  receiving  end  we  have  the  following  conditions: 
Line  voltage  30  kv.  between  wire  and  neutral,  load  current  at  re- 
ceiving end  150  amperes  at  90  per  cent  power  factor,  lagging  cur- 
rent. Phase  angle  lagging  25°  50'. 

Calculating  the  inductance  and  capacity  of  the  line  by  formulae 
given  in  Chapters  II  and  III  we  find, 

r  =  0.33  ohm  per  mile, 
L  =  2.13  mh.  per  mile, 
C  =  0.014  mf  .  per  mile. 

The  total  resistance,  inductance  and  capacity  of  the  line  are 
16.5  ohms,  0.1065  henry  and  0.7  mf.  respectively. 
The  electrical  conditions  at  the  receiving  end  are 

Er  =  30  kv., 

Ir  =  I/  -  jlr"  =  135  -  j  65.4  amp. 

In  formula  (68)  we  neglect  the  capacity  of  the  line  entirely,  hence 
introducing  the  values  of  r,  L  and  w  we  get 
Ig=Ir  =  135  -  j  65.4  amp., 
Eg  =  530  +  2.63  +  2.23J  +  j  {5.42  -  1.08J 

=  34.86  +.7  4.34. 
The  absolute  value  of  e.m.f.  at  generator  is 

Eg  =  vX/2  +  Eg"*  =  35.1  kv. 

The  voltage  at  generator  leads  the  voltage  at  receiver  by  an 
angle 


hence  the  current  at  generator  end  lags  behind  the  generator  e.m.f. 
by  the  angle  25°  50'  +  7°  5'  =  32°  55'. 
The  power  factor  at  the  generator  end  is 

cos  (32°  55')  =  0.84. 
The  efficiency  of  transmission 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    153 


Example.  —  The  same  conditions  as  in  above  example  except  that 
the  line  is  150  miles  long. 
We  have  as  in  the  previous  example, 

Ig=L  =  135  -  j  (65.4)  amp., 
Eg  -  44.58 +j  13.02  kv. 

The  absolute  value  of  e.m.f.  at  generator  is 

Eg  =  V(44.58)2  +  (13.02)2  =  46.4  kv. 

The  generator  e.m.f.  leads  the  voltage  at  the  receiving  end  by 
the  angle 


The  current  at  generator  end  lags  behind  Eg  by  the  angle  25°  50' 
+  16°  16'  =  42°  6'. 

The  power  factor  at  generator  end  =  cos  (42°  6')  =  0.742. 

The  efficiency  of  transmission  is 
30  X  150  X  0.9 


46.4X150X0.742 


=  7SA  per  Cent' 


L       r 
J                           ~2~'T 

=^X 

, 

2    '    2 

T  *•              —  1 

J   T* 

•^7                     —  I 

C 

Vr.I, 

7 

\  \ 

. 

. 

FIG.  46. 

Case  II.  Capacity  of  the  line  shunted  across  the  middle  of  the 
line. 

If  the  admittance  of  receiving  circuit  g  ±  jb  and  the  voltage 
Er  are  given,  we  have  Ir  =  Er  (g±jb)  =  Ir'  db  jlr". 

The  voltage  and  current  at  the  generator  end  are  given  by 
the  formulae 

Ig  =  \KIr'^\CurIr"\  +Jl±CurIr'±KIr"+ErCul  =lg'±jlg",  (70) 
Eg  =  Er  +  ;jr  (//  +  //)  -  JLu  (/,"  ±  //Of 

+  j  \  i  r  (//'  ±  Irff)  +  i  Leo  (//  +  //)  \  =  Eg'±  jEg",     (71) 
where  K  =  1  -  J 


154  FORMULAE  AND  TABLES  FOR  THE 

The  double  signs  in  the  terms  of  equations  (70)  and  (71)  have 
the  same  significance  as  in  equations  (68)  and  (69).  The  upper 
sign  is  to  be  used  when  Ir"  is  positive,  that  is,  the  receiving  current 
leading  voltage  at  receiving  end,  and  the  lower  sign  is  to  be  used 
when  Ir"  is  negative. 

If  the  voltage  and  current  are  given  at  the  generator  end,  that 
is,  Eg  and  Ig  =  IJ  ±  jlg"  are  given,  the  current  and  voltage  at 
the  receiving  end  are  given  by  the  formulae: 


Ir  =  I  KI=F  J  rCuIg"\  +  j\±  KIg" 

=  lr'±jlr",  (72) 

ET  =  Eg  -  \\r  (7/  +  //)-  JTxo  (/,"  db  7/0} 

-  JlMIr"  ±  /,")  +  i  Txo  (/„'+//)  J  =Eg'±jE,»,       (73) 
where  as  before  K  =  1  —  \  CT/co2. 

As  an  illustration  and  for  comparison  we  shall  work  out  ex- 
amples given  above  by  formulae  (70)  and  (71). 

r  =  16.5  ohms,     L  =  0.1065  henry,     C  =  0.7  mf. 

K  =  0.995,     Ceo  =  264  X  1Q-6. 
Er  =  30  kv.,     Ir  =  135  -  j  65.4  amp. 

The  phase  angle  at  the  receiving  end  is  25°  51'  lagging. 

7,  H  134.3  +  0.14J  +  j  ;o.29  -65.07  +7.92  j  =  134.44  -j  56.86  amp.; 


absolute  value  of  Ig  =  V  (134.44)2  +  (56.80)2  =  146  amp. 
Eg  =  (30  +  2.22  +  2.45)  +  j  (-1.01  +  5.41)  =  34.67  +  j 4.4kv; 

absolute  value, 

Eg  =  V  (34.67)2  +  (4.4)2  =  34.95  kv. 

r£>   or> 

The  phase  angle  between  Ig  and  Er  is  tan-1     '   .  =  32°  16r,  lagging. 

1O4.4 

4  4 

The  phase  angle  between  Eg  and  Er  is  tan-1  =  7°  15',  leading. 

o4:.b7 

Ig  lags  behind  Eg  by  the  angle  23°  16r  +  7°  15r  =  30°  31'. 
Power  factor  at  generating  end  =  cos  (30°  31')  =  0.862. 
Efficiency  of  transmission, 

150  X  30  X  0.9 


146X34.95X0.862 


=  92.1  percent. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    155 


Example.  —  The  same  conditions  as  in  the  example  above  except 
that  the  length  of  the  line  is  150  miles  long. 

Er  =  30  kv.,    Ir  =  135  -  j  65.4  amp.,    K  =  0.952. 
Ig  =(128.52+1.26)  +  j(2.61-62.26+23.76)  =  129.8  -  j  35.9  amp. 

Absolute  value  of  current  at  receiving  end, 

Ig  =  V(129.8)2  +  (35.9)2  =  135  amp. 
Eg  =(30  +  6.55+6.1)  +j(-  2.51  + 15.95)  =42.65  +j  13.44  kv. 

Absolute  value  of  voltage, 

Eg  =  V  (42.65)2  +  (13.44)2  =  44.7  kv. 

35  9 

The  phase  angle  between  Ig  and  Er  is  tan"1     „     =  15°  30'  lagging. 

The  phase  angle  between  Eg  and  Er  is 

10  A  A 

^^^iWr  leading. 

Ig  lags  behind  Eg  by  the  angle  15°  30'  +  17°  30'  =  33°. 
Power  factor  at  generating  end,  cos  (33°)  =  0.84. 

Efficiency  of  transmission 

150  X  30  X  0.9 
11  =  135  X  44.7  X  0.84  =  =  8°  per 

Lr 


E, 


h 


FIG.  47. 

Case  III.  Half  of  line  capacity  shunted  across  at  each  end  of  the 
line. 

As  in  previous  cases  /  and  II  we  shall  assume  that  Er  and  /,. 
are  given  in  magnitude  as  well  as  phase  relation,  that  is, 

/r=I/+j(±/r").  (74) 


156  FORMULAE  AND  TABLES  FOR  THE 

The  values  of  the  voltage  and  current  at  the  generator  end  are 
given  by  the  formulae 


Eg  =  \ErK  +  rlr'  -  Lulr"l  +j  li  CW  Er  +  rlr"  +  Leo//} 

=  #,'+j(±tf,")  (75) 

Ig  =  5//-pV'C<oi  +j  \  i  #rCco+PVCo>+7r"J  =lg'±jla".  (76) 

The  values  of  Ir"  and  Eg"  are  to  be  taken  positive  or  negative 
depending  on  the  sign  of  these  quantities  in  the  parentheses  in  (74) 
and  (75). 

If  Eg  and  Ig  are  given  in  magnitude  as  well  as  in  phase  relation, 

Ig  =Ig'+j(±Ia"}.  (77) 

We  have  the  following  expressions  for  Er  and  Ir: 
Er  =  \EgK  -  rlgf  +  Lco//'J  +j\$Ea  Cur  -  Leo//  -  rla"\ 

=  Er'+j(±Er"),  (78) 

Ir=\I0'+$Er"C<*]+j\Ig"-$EaC<*-$Er'C<*}  =  //+j(zb//;),  (79) 

where 

K  =  l-\LCu\ 

In  the  following  two  examples  we  shall  use  the  same  data  as  in 
examples  considered  under  Case  II. 
Example.  —  Fifty-mile  transmission  line. 

r  =  16.5  ohms., 
L  =  0.1005, 
C  =  0.7  mf., 
K  =  0.995, 

Er  =  30  kv.,     Ir  =  135  -  j  65.4  amp. 
Eg  =  {  29.85  +  2.23  +  2.63  {  +  j  {0.065  -  1.08  +  5.42  j 
=  34.7  +j  4.4  kv. 

Absolute  value  of  voltage  is 

Eg  =  V(34.7)2  +  (4.4)2  =  35  kv. 
Ig  =  \  135-0.58  l+j  [3.96+4.58-  65.4  j  =  134.4  -j  56.86  amp. 

Absolute  value  of  current  is  Ig  =  V(134.4)2  +  (56.S6)2  =  145.9 
amp. 

Eg  leads  Er  by  the  angle  tan-1  7        =  7°  14', 


Ig  leads  Er  by  the  angle  tan-1  =  22°  57', 

Ig  lags  behind  Eg  by  the  angle,  22°  57'  +  7°  14'  =  30°  11'. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS     157 

Power  factor  at  generating  end, 

cos  (30°  110  =  0.864. 
Efficiency  of  transmission, 

150  X  30  X  0.9 
12  =  145.9  X  35  X  0.864  =  9L8  per  Cent' 

Example.  —  One  hundred  fifty-mile  three-phase  transmission  line. 

R  =  49.5  ohms, 
L  =  0.3195, 
C  =  0.21  mf., 
K  =  0.952. 

Er  =  30  kv.,       Ir  =  135  -  j  65.4  amp. 
.      Eg  =  S28.5  +  6.7  +  7.9J  +  j  {0.6  -  3.25  +  16.25J 
=  43.1  +    13.6  kv. 


The  absolute  value  of  the  generator  voltage, 


Eg  =  V  (43.1)2  +  (13.6)2  =  45.2  kv. 
Ig  =  J135  -  5.4}  +j  {11.9  +  17.05  -  65.4J  =  129.6  -  j  36.5  amp. 

Absolute  value  of  current, 

Ig  =  V(129.6)2  +  (36.5)2  =  134.7  amp. 
Eg  leads  Er  by  the  angle  tan-1^  =  17°  30', 

Ig  lags  behind  Er  by  the  angle  tan  ~1^-a  =  15°  45', 

izy.o 

Ig  lags  behind  Eg  by  the  angle  17°  30'  +  15°  45'  =  33°  15'. 
The  power  factor  at  generator  end, 

cos  (33°  15')  =  0.836. 
The  efficiency  of  transmission, 
150  X  30  X  0.9 


134.7  X  45.2  X  0.836 


=  79'5  per  Cent' 


158 


FORMULAE  AND  TABLES  FOR  THE 


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CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    159 


TRANSFORMERS 

Any  two  electrical  circuits  which  are  magnetically  interlinked 
so  that  when  a  current  flows  in  one  circuit  it  will  induce  an  e.m.f. 
in  the  other  circuit  constitutes  a  transformer.  For  all  commer- 
cial work  the  iron-core  transformer  is  used.  In  high-frequency 
work,  as  in  the  case  of  wireless  telegraphy  for  instance,  air-core 
transformers  are  used  exclusively. 

We  shall  first  give  the  equation  governing  the  action  of  an  air- 
core  transformer,  and  then  give  a  brief  discussion  of  the  theory 
of  the  iron-core  transformer. 

Air-core  Transformer.* — Denote  by  LI,  Ri  and  L2,  R2  the 
total  inductance  and  resistance  of  the  primary  and  secondary  cir- 
cuit respectively  and  by  M  the  mutual  inductance  between  the 
two  circuits.  We  shall  also  assume  an  e.m.f.  of  sine  wave  applied 
to  the  primary  circuit. 

The  equations  expressing  the  reactions  in  the  transformer 
circuits  are 


dt 


dt 


(80) 


from  which  we  obtain  the  values  of  the  currents  in  the  two  cir- 
cuits as  follows : 


(81) 


*  Alex.  Russell,  "Theory  of  Alternating  Currents,"  Vol.  1,  Chap.  X. 
Clerk  Maxwell,  "  A  Dynamical  Theory  of  the  Electromagnetic  Field,"  Phil. 
Trans.,  1865,  p.  475. 


160  FORMULAE  AND  TABLES  FOR  THE 


Ri  + 

*  ~  7)  (82) 


2  /*  2  7    2    2  _1_   7?  2 

=  — jg — *     22     ==  -L/2  CO     -j-  /t/2  . 

It2 

The  value  of  the  current  in  the  primary  circuit  is  the  same  as 

MVR2 

that  in  a  simple  circuit  whose  resistance  is  Ri  -\ ^ —  and  in- 

Tl  yC0         ft  T 

ductance  is  LI ^— -,  that  is,  the  secondary  circuit  influences 

the  primary  circuit  in  the  same  way  as  if  its  resistance  were 
increased  and  the  inductance  decreased.      In  other  words  the 

effective  resistance  in  R\  +  • ^  instead  of  Ri  and  effective 

inductance  is  LI ^-^  instead  of  LI.     If  we  neglect  the  re- 

£2 

sistance  of  the  secondary  circuit,  the  effective  inductance  of  the 
primary  circuit  is 

7l/f2/,27"  /  M  2  \ 

(83) 


The  quantity  cr  is  called  the  leakage  factor. 

Resonance  Transformer.*  —  If  we  introduce  capacities  in 
the  primary  and  secondary  circuits,  the  expressions  for  the  cur- 
rents have  exactly  the  same  form  as  that  given  by  equations 
(81)  and  (82)  except  that  LICO  and  L2co  are  replaced  by  the  terms 

LICO  —  -^ —  and  L2co  —  ^ —  where  Ci  and  C2  are  the  capacities  in 
C  ico  C  2co 

primary  and  secondary  circuits  respectively.     Now  suppose  the 
primary  and  secondary  circuits  are  separately  tuned  so  as  to  be  in 

*  G.  W.  Pierce,  "Theory  of  Electrical  Oscillations  in  Coupled  Circuits," 
Proc.  Am.  Acad.,  Vol.  46,  pp.  293-322,  Jan.,  1911. 

J.  S.  Stone,  "The  Maximum  Current  in  the  Secondary  of  a  Transformer," 
Phys.  Rev.,  Vol.  32,  pp.  398-405,  April,  1911. 

G.  Benischke,  "Der  Resonanztransformotor,"  E.  T.  Z.,  1907,  p.  25. 

J.  Bethenod,  "tiber  den  Resonanztransformotor,"  Jahrbuch  der  Drahtlosen 
Telegraphic,  Vol.  I,  pp.  534-570. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    161 
resonance  for  the  frequency  of  the  impressed  e.m.f.     Then 

-ei  =  o,    z°  =  «"        (84) 


(85) 


and  equations  (81)  and  (82)  reduce  to 
,         ER%  cos  a)t 


A  -f  M2co 
EM  co  sin  <*>t 


1  2=  Current  in  Secondary 

t-t  to  Co  £*  en  cr<  -3  ooto  c 

/ 

^3- 

^ 

^ 

/ 

\ 

^ 

fe 

/ 

\ 

^ 

1 

'•^ 

\ 

^>^_ 

"^ 

•-^. 

•^~«. 

1 

Current  variation  in  the  Secondary 
of  a  resonance  Transformer 
R  i  =  20     Ohms 
R  2  =12.5       » 

\ 

"^^ 

/ 

/ 

/ 

1 

Mu 

012 


4      56      7      8      9     10    11     12    13    14     15    16    17     18    19    20 


FIG.  49.  —  Current  variation  in  the  secondary  of  a  resonance  transformer  for 
different  degrees  of  coupling. 

The  current  in  the  secondary  circuit  is  a  maximum  when  MV 
=  RiR2  and 

E  sin  at  ,     . 


If  we  have  only  a  condenser  in  the  secondary  as  shown  in 
Fig.  50,  we  have  the  following  expressions  for  the  currents  in  the 
primary  and  secondary  circuits  and  the  voltage  on  the  condenser. 


E 


V  I  C  1      2J          CO    j      x 


(87) 


162 


FORMULAE  AND  TABLES  FOR  THE 

E  Mu  sin  (ut  —  <j>) 


FIG.  50. 

The  voltage  on  condenser  is  given  by 

n'M 


V/{(M'- 


(88) 


(89) 


The  condition  for  maximum  currents  in  the  circuits,  that  is,  the 
resonance  condition,  is 


or 


l-£ 


=  0. 


CL2  (1  - 


,  where  .K2  = 


M 


(90) 


If  the  constants  of  the  circuits  are  fixed  so  as  to  satisfy  equation 
(90),  and  if  we  neglect  RiR2  which  is  generally  small,  we  have  the 
following  expressions  for  the  currents  and  voltage  in  the  circuits: 


—  EMu  cos  cot 


sin  (<at 


cos  (at, 


(91) 


(92) 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    163 
—  EM  sin  ut 


Ceo 


(93) 


]L 

Ceo 


The  form  of  transformer  discussed  above  is  used  extensively  in 
the  production  of  high-potential  high-frequency  currents.  The 
arrangement  of  circuits  used  is  indicated  in  Fig.  51. 


FIG.  51. 

E  is  an  alternator,  M}  a  high  potential  transformer. 

When  the  condenser  C  is  charged  to  its  maximum,  a  spark  occurs 
at  the  gap  g  which  short-circuits  the  condenser  and  an  oscillatory 
discharge  takes  place  in  the  circuit  gCL0.  The  period  of  the 
oscillations  depends  upon  the  constants  of  the  circuit. 

The  equations  given  above  refer  only  to  the  permanent  condi- 
tions, but  when  the  condenser  charges  and  discharges  at  very 
frequent  intervals,  the  perma- 
nent conditions  may  not  be 
even  reached  and  the  transient 
phenomena  becomes  of  great 
importance.  We  shall  give  the 
formulae  for  the  transient  terms 
in  the  next  chapter. 

Another  form  of  transformer 
which  finds  considerable  appli-  JTIG  52. 

cation  in  a  high-frequency  work 

is  one  in  which  there  is  only  a  condenser  in  the  primary  circuit  as 
shown  in  Fig.  52.  This  arrangement  corresponds  to  a  receiving 
circuit  in  radiotelegraphy  having  an  untuned  secondary. 


164  FORMULAE  AND  TABLES  FOR  THE 

If  we  assume  that  the  incoming  signals  acting  on  the  antenna 
are  undamped,  then  we  have  the  following  expressions  for  the 
amplitudes  of  the  currents  hi  the  primary  and  secondary  circuits. 


E  VL22o>2 


V  J 


M2co2  +  #ifl2- 

\  ^1W/J  [        \  UiCO/ 

(94) 


(95) 

If  we  adjust  the  inductance  and  capacity  of  the  primary  circuit 
so  as  to  be  in  resonance  for  the  frequency  of  the  e.m.f.  impressed 

on  the  circuit,  that  is,  if  we  make  L\u  —  7^—  =  0,  equations  (94) 


and  (95)  reduce  to 


E  VL2 


=  EM* 


2  +  R^R^l  + 

The  coupling  or  the  mutual  inductance  which  gives  the  maximum 
current  in  the  secondary  circuit  is  determined  by  the  equation 


+  L22o>2  =  BA.  (97) 

For  the  value  of  Mco,  given  by  equation  (97),  we  have  for  the 
currents  in  the  primary  and  secondary  circuits 

EZ* 


l  V2  (Z22  +  R2Z2)  '  R1 


As  an  illustration  we  may  use  the  following  constants: 

#1  ='  15  ohms,     R2  =  200  ohms,     LI  =  1  mh.,     L2  =  2  mh., 
C  =  0.001  mf.,      co  =  106. 

By  formula  (97)  the  value  of  M  which  makes  72  a  maximum  is 

M  =  0.1736  mh. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    165 

Introducing  these  constants  in  equations  (96)  we  get  the  fol- 
lowing results: 


M. 

fc 

/2- 

mh. 

0.1 
0.15 
0.1736 
0.25 
0.50 
1.00 

6.1  XlO-2# 
5.1  XlO-2# 
4.48X10~2# 
2.80XlO-2# 
0.78xlO-2# 
0.20XlO-2# 

3.2  X10~3# 
3.84xlO-3# 
3.89XlO-3# 
3.47xlO-3# 
1.96XlO-3# 

i.ooxio-3# 

The  Iron-core  Transformer.*  —  In  the  iron-core  transformer 
we  are  chiefly  concerned  with  efficiency  ratio,  regulation  and 
exciting  current.  It  is  preferable  to  consider  the  transformer 
from  the  point  of  view  of  the  leakage  inductance,  as  this  method 
leads  to  simpler  formulae  than  does  the  treatment  from  the 
mutual  inductance  point  of  view,  and  at  the  same  time  brings  out 
more  clearly  the  relations  between  the  various  factors  in  the 
transformer. 

The  general  design  formula  is 

E  =  irVZnQN  X  10~8,  (99) 

where  E  is  the  effective  voltage  induced  in  a  winding,  n  the  fre- 
quency, 3>  the  maximum  total  flux  linking  with  the  winding  and 
N  the  number  of  turns. 

Example.  —  In  a  60-cycle  transformer  whose  core  is  10  cm. 
square  and  which  is  worked  at  10,000  lines  per  centimeter,  the 
volts  per  turn  are 

E  =  7rV2  X  60  X  102  X  10,000  X  10~8  =  2.7. 

Let  Ri,  R2  =  primary  and  secondary  resistance  of  windings, 

Xij  X2  =  primary  and  secondary  leakage  reactances  of  windings, 
Zi,  Z2  =  primary  and  secondary  internal  impedances, 
/o  =  exciting  current, 
I  =  load  current  measured  on  primary  side, 

*  William  Cramp  and  C.  F.  Smith,  "Vector  and  Vector  Diagrams  Applied 
to  the  Alternating  Current  Circuit,"  Ch.  V. 

C.  P.  Steinmetz,  "Theory  and  Calculation  of  Alternating  Current  Phe- 
nomena," 3d  edition,  C.  XIV  and  XV. 

V.  Karapetoff,  "The  Electric  Circuit." 


166         .        FORMULAE  AND  TABLES  FOR  THE 

EI,  E2  =  primary  and  secondary  e.m.f.'s, 
E2,o  =  secondary  e.m.f  .  at  no  load, 

6  =  phase  angle  of  connected  load,  being  positive  when 

the  secondary  voltage  leads  the  secondary  cur- 
rent, cos  6  being  the  power  factor  of  the  load, 

7  =  angle  by  which  EI  leads  the  exciting  current, 
a  =  angle  between  EI  and  E2  reversed, 

K  =  ratio  of  turns,  primary  to  secondary. 

Mathematically  it  is  convenient  to  refer  the  impedances  to  the 
primary.  This  may  be  done  since  an  impedance  Z2  in  the  second- 
ary is  equivalent  to  an  impedance  K2Z2  in  the  primary.  Hence 
we  may  write 

R  =  R!  +  KZR2, 


K*X 


(100) 
Z  =  Zl 

In  computing  the  efficiency  of  ordinary  power  transformers  it 
is  usual  to  neglect  the  copper  loss  due  to  the  exciting  current 
and,  unless  the  contrary  is  distinctly  stated,  the  efficiency  refers 
to  unity  power  factor  load.  Hence  the  formula  is  simply, 

output  KE2I  .  .  _  .  . 

per  cent  efficiency  =  -*—  •—  =  ^^  T   ,    T9P   ,  -  -,  --     (101) 
input       KE2I  +  PR  +  core  loss 

The  core  loss  is  practically  equivalent  to  the  power  consumption 
on  open  secondary. 

In  an  ideally  perfect  transformer  EI  would  be  equal  to  KE2, 
but  in  the  actual  transformer  these  differ  by  the  impedance 
drops  in  the  windings  due  to  the  load  current  and  to  the  ex- 
citing current,  or 

&  -  KE2  =  (70  +  7)  Zx  +  KZIZ2  =  70Zi  +  7Z,        (102) 

where  the  dotted  capitals  indicate  that  the  quantities  are  vector 
quantities.  This  is  the  general  equation  for  every  form  of  trans- 
former. 

The  leakage  reactance  of  the  windings  is  ordinarily  determined 
by  the  '  'short-circuit"  test,  in  which  one  winding  is  short-circuited 
and  rated  current  passed  through  the  windings  by  applying  a 
voltage  to  the  other  winding.  From  the  value  of  this  voltage,  the 
power  required  and  the  current,  the  resistance  R  and  the  reactance 
X,  equivalent  to  a  simple  impedance,  may  be  readily  computed. 

In  the  derivation  of  the  formulae  for  regulation,  ratio  and 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    167 

phase  angle,  slight  approximations  are  necessary  to  put  the 
formulae  in  usable  form,  but  in  all  practical  cases  the  formulae 
are  more  accurate  than  the  measurements  of  the  quantities  in- 
volved can  be  made  experimentally.*  The  ratio  of  terminal  volt- 
ages at  any  load  is 

Ei  _  K  i  IRcosO  +  IXsmO      (IR sin 6  -  IX cos 0) 
#2~KH  E2 

.  IQ  (Ri  cos  7  +  Xi  sin  7)  t 

T? ~~  > 

J&2 

for  non-inductive  load, 

E,_^  ,  IR 

~        {" 


E* 

and  for  no  load, 

EI  _  „      ZQ  QRi  cos  y  -j-  Xi  sin  7) 

The  regulation  at  any  power  factor  expressed  in  per  cent  is 

per  cent  regulation  =  100(    2>°  r, — - ) 

\    A^2     / 

"sin 0      (IRsmd  -  IXcosd)2 

for  non-inductive  load, 

f  IR         I2X*    1 
per  cent  regulation  =  100  i  ^^-  -f-  ^  ^2^  2  X . 

^  /VIZ/2          ^  IV  H/2    J 

The  phase  angle  of  the  transformer  is  given  by 

sin  a  =^  \Xcosd  —  Rsmdl  +  ^-  jXiCOS7  —  i2isin7j ;  (108) 
and  for  the  non-inductive  load, 

sina  =-^-+-^(^Licos7  —  ^isin7),  (109) 

and  for  no  load, 

sin  a0  =  j=r  (X  cos  7  —  Ri  sin  7).  (110) 

*  A  derivation  of  these  formulae  showing  the  magnitude  of  the  approxima- 
tion involved  will  appear  in  Vol.  10  of  the  Bulletin  of  the  Bureau  of  Standards, 
by  P.  G.  Agnew  and  F.  B.  Silsbee. 


168  FORMULAE  AND  TABLES  FOR  THE 

The  reversed  secondary  voltage  leads  the  primary  voltage  for  in- 
ductive loads  and  ordinarily  at  no  load,  but  lags  behind  it  for 
non-inductive  loads. 
Example.  —  Consider  a  2  kva.,  Wir-volt,  60-cycle  transformer. 

Let    #1  =  6,    #2  =  0.065,     X  =  15,     70  =  0.12,     cos  7  =  0.4, 
K  =  10,     7  =  2,     R  =  R!  +  K*R2  =  12.5. 

If  we  assume  the  reactance  drop  to  be  divided  equally  between 
primary  and  secondary,  Xi  =  7.5. 

The  computations  are  correct  to  0.01  per  cent,  although  the 
measurements  are  not. 

Then  by  (105)  the  no-load  ratio  is 

&  f  1+  0.12  (6X0.4  +  7.5X0.92)1 

#2,o  V  100°  J 

The  regulation  at  0.6  power  factor,  lagging  current,  is  by  (106) 
25X0.6+30X0.8  ,   (25X0.8-30X0.6)2 


ft 

100 


[ 

t 


2x100x10,000 


For  leading  current  the  terms  in  (106)  which  contain  sin  0  be- 
come negative  so  that  at  the  same  power  factor  leading  current 
the  regulation  is 

100  (-  0.0090  +  0.0007)  =  -  0.83  per  cent. 

The  minus  sign  shows  that  under  this  condition  the  secondary 
voltage  is  increased  by  the  leading  load  current. 
At  non-inductive  load  the  regulation  is  by  (107) 

(    25  900      ) 

100  1000  ~  =  *'55  per  cent' 


The  phase  angle  at  0.6  power  factor,  lagging  current  is  by  (108) 
30  X  0.6  -  25  X  0.8  ,  0.12(7.5X0.4-6X0.92) 

-Tooo-  ~looo~  ~a°023' 

a  =-8', 

the  minus  sign  meaning,  according  to  the  convention  adopted, 
that  the  reversed  secondary  voltage  leads  the  primary  voltage. 
From  formulae  (109)  and  (110)  we  have  for  non-inductive  load, 

3  a     .  0.12  (7.5  X  0.4  -  6  X  0.92)  0      , 

1000  +  -  ~          -  =  a°297'    a  =  l  42  ' 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    169 

the  reversed  secondary  voltage  lagging;  and  for  no  load 

0.  12  (7.5  X  0.4  -  6  X  0.92  , 

sm  a  =  -  10QQ  -  =  -  0.0003,     a  =  1  , 

the  reversed  secondary  voltage  leading. 

The  Current  Transformer.*  —  While  the  foregoing  relations 
hold  also  for  the  current  transformer,  the  conditions  of  use  and 
the  relations  desired  are  so  different  that  a  different  method  of 
treatment  is  far  more  convenient.  The  resistance  and  reactance 
connected  to  the  secondary  are  kept  as  low  as  possible,  and  the 
iron  is  worked  at  low  flux  densities,  the  density  being  practically 
proportional  to  the  current.  The  ratio  and  phase  angle  of  the 
current  transformer  are  independent  of  the  resistance  and  react- 
ance of  the  primary. 
Let  1  1,  1  2  =  primary  and  secondary  currents, 

K  =  ratio  of  turns,  secondary  to  primary, 
M  =  wattless  component  of  exciting  current, 

F  =  core  loss  component  of  exciting  current, 

i  total  secondary  reactance 
0  =  tan-1—  —  j—  —  T-T     -j 

total  secondary  resistance 

6  =  angle  between  /i  and  72. 
The  ratio  of  transformation  is 

Jt  _          M  sin  0  +  F  cos  <f> 


and  the  phase  angle  is  given  by 

(112) 


If  the  load  connected  to  the  secondary  circuit  is  non-inductive 
and  if,  as  is  usually  the  case  in  high-grade  transformers,  the 
leakage  reactance  of  the  secondary  winding  may  be  neglected, 
equations  (111)  and  (112)  become 

fx=X  +  f,  (113) 

Iz  *2 

tan  0=|^.  (114) 

A./2 

Equations  (111)  to  (114)  neglect  second-order  terms,  but  are  suffi- 
ciently accurate  for  all  practical  work. 

*  P.  G.  Agnew,  Bull.  Bureau  of  Standards,  7,  431,  1911.     Reprint  No.  164. 


170  FORMULAE  AND  TABLES  FOR  THE 

Example.  —  A  1000  to  5  ampere  current  transformer  has  198 
secondary  turns  and  one  primary  turn  and,  when  the  total  re- 
sistance and  the  total  reactance  of  the  secondary  circuit  are  0.4 
and  0.5  ohm  respectively,  the  core  loss  is  0.02  watt,  and  the  ex- 
citing current  is  8  amperes  for  a  secondary  current  of  2.5  amperes. 
Then  K  =  198, 

sin  0  =  =  =  0.78,     cos  <£  =  0.62. 

VO.16  +  0.25 


The  voltage  induced  in  the  primary  is  TJ¥  2.5  VO.16+0.25  =  0.0081 . 
Hence  F,  the  power  component  of  the  exciting  current,  is 

0.02  -f-  0.0081  =  2.47  amp. 

and  M,  the  wattless  component,  is  \/82  —  (2.47) 2  =  7.6  amp.  Sub- 
stituting these  values  of  F}  M  and  <£  in  equations  (111)  and  (112), 
we  have  for  the  ratio  and  phase  angle, 

/!  .   7.6X0.78  +  2.5X0.62 

7-  =  iyo  ~r  •  o  c 

12  ^.O 

and 


the  reversed  secondary  current  leading  the  primary  current. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    171 


CHAPTER  V 
TRANSIENT   PHENOMENA 

IF  the  electrical  conditions  of  a  circuit  are  disturbed  in  any  way, 
as  for  instance  by  a  change  in  the  electrical  constants  of  the  cir- 
cuit, or  a  change  in  the  electromotive  force  acting  on  the  circuit,  a 
readjustment  of  the  current  and  potential  in  the  circuit  will 
necessarily  follow.  The  permanent  state,  however,  is  not  reached 
instantaneously;  it  requires  an  appreciable  time  interval  before 
electrical  equilibrium  is  again  established.  The  electric  phe- 
nomena which  occur  in  the  time  interval  before  the  permanent 
state  is  reached  again  have  been  properly  designated  Transient 
Electric  Phenomena.  These  phenomena  are  of  frequent  occur- 
rence and  of  considerable  importance  in  electrical  engineering, 
particularly  whenever  high  frequencies  or  high  potentials  are 
employed. 

It  is  not  within  the  scope  of  this  book  to  enter  into  any  discus- 
sion of  the  physical  interpretation  of  these  phenomena  but,  as  in 
the  previous  chapters,  we  shall  state  only  the  problem,  that  is,  the 
electrical  conditions  governing  the  production  of  these  phenomena, 
and  give  the  mathematical  formulae  expressing  the  distribution 
of  the  current  and  potential  in  the  circuit.  Wherever  desirable 
we  shall  illustrate  the  formulae  by  numerical  examples  and  curves. 

In  this  chapter  we  shall  limit  ourselves  to  a  consideration  of 
circuits  of  localized  inductance  and  capacity.  The  transient 
phenomena  which  occur  in  a  circuit  of  distributed  inductance  and 
capacity  will  be  discussed  in  Chapter  VI. 

INDUCTANCE  AND  RESISTANCE  IN  DIRECT  CURRENT  CIRCUITS* 

If  we  close  suddenly  a  direct  current  circuit  containing  resist- 
ance and  inductance  the  current  does  not  assume  its  permanent 
value  instantaneously.  It  requires  a  certain  interval  of  time  to 

*  Ch.  P.  Steinmetz,  "Theory  and  Calculation  of  Transient  Electric  Phe- 
nomena and  Oscillations,"  Ch.  III. 

J.  L.  LaCour  and  O.  S.  Bragstad,  "Theorie  der  Wechselstrome,"  Ch. 
XXIV. 


172  FORMULAE  AND  TABLES  FOR  THE 

build  up  the  magnetic  field  in  the  inductance  coil,  and  during  that 
interval  there  is  a  back  e.m.f.  produced  by  the  inductance  which 
prevents  the  current  from  rising  immediately  to  its  permanent 
value.  The  value  of  the  current  at  any  instant  of  time  after 
closing  the  circuit  is  given  by 

I  =  7.(l  -«*'),  (1) 

Tp 

where  I,  =  •=    is  the  value  of  the  current  in  the  steady  state, 
it 

and  the  time  is  counted  from  the  instant  of  closing  the  circuit. 

At  t  =  0,  7  =  0, 

t  —  oo  ,          I  =  1  8,  permanent  value 

-5  is  called  the  time  constant  of  the  circuit,  and  its  reciprocal 
K 

T) 

value  a  =  -j-  is  called  the  damping  factor  of  the  circuit.     The 
Li 

time  required  for  the  current  to  rise  to  n  per  cent  of  its  steady 
value  is 


100 

The  smaller  the  inductance  and  the  larger  the  resistance  the 
quicker  will  the  current  attain  its  permanent  value.  In  a  per- 
fectly non-inductive  circuit  the  current  would  assume  its  perma- 
nent value  instantaneously. 

Example.  —  When  L  =  0.  1  henry,  R  =  5  ohms,  what  is  the  time 
required  for  the  current  to  rise  to  90  per  cent  of  its  permanent 
value?  By  equation  (2), 

T  =  ^  log,       *       =  0.02  X  2.3  =  0.046  sec. 
o        i  —  u.y 

If  in  the  above  example  we  make  L  =  1  henry,  then  the  time  re- 
quired for  the  current  to  rise  to  90  per  cent  of  its  permanent  value 
is  0.46  second,  ten  times  as  large  as  in  the  former  case.  In  0.046 
second  the  current  would  in  this  case  rise  only  to  21  per  cent  of 
its  permanent  value. 

On  removing  the  e.m.f.  from  a  direct  current  circuit,  the  cur- 
rent does  not  vanish  immediately.    The  magnetic  field  of  the 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    173 


inductance  produces  an  e.m.f.  which  causes  a  current  I  to  flow  in 
the  circuit  the  value  of  which  is  given  by  the  expression 

_R 
I  =  I.e    l  ,  (3) 


1.0 

0.9 
0.8 
0.7 
0.6 

0.4 
0.3 
0.2 
0.1 

—  —  • 

_    - 

_—  —  • 

- 

—  — 



1  

^ 

^ 

—  - 

^ 

^ 

, 

/ 

/ 

/ 

Rise  of  Current  in  Inductive  Circuit 

/ 

/ 

#=10  Ohms,  L=  0.5  Henry 

/ 

/ 

/ 

0.01  0.02  0.03  0.04  0.05  0.06  0.07  0.08  0.09  0.10  0.11 0.12  0.13  0,14  0.15  0.16  0.17  0.18  0.19  .20 

Time  in  Seconds 
FIG.  53. 

where  I8  is  the  steady  value  of  the  current  before  the  e.m.f.  is 
removed.  The  e.m.f.  generated  by  the  magnetic  field  of  the  in- 
ductance coil  is 

a          R 


F-r 
E~Ldi 


(4) 


1.00 


\ 


Decay  of  Current  in 

Inductive  Circuits 

B=10  Ohms  L=0.5  Henry 


0.01  O.OS  0.03  0.04  0.05  0.06  0.07  0.08  0.09  0.10  0.11 0.12  0.13  0.14  0.15  0.16  0.17  0.18  0.19.20 
Time  in  Seconds 
FIG.  54. 

In  Figs.  53  and  54  two  curves  are  given  showing  the  rise  and 
decay  of  the  currents  in  an  inductive  circuit  on  closing  and  open- 
ing the  circuit. 


174  FORMULAE  AND  TABLES  FOR  THE 

CLOSING  AN  ALTERNATING  CURRENT  INDUCTIVE  CIRCUIT 

Let  us  assume  that  at  the  instant  of  closing  the  circuit  the 
e.m.f.  of  the  generator  is 

e  =  E  sin  (ut  +  ^)  =  E  sin  ^=0.  (5) 

The  current  in  the  permanent  state  is 

/'  =  /  sin  (at  +  $  -  0)  =  /  sin  (f  -  0)f  =0,  (6) 

,  E  wL 

where  /  =     ,  =  >      tan     =  ~ 

W2L2 


The  current,  however,  will  not  rise  instantly  to  its  permanent 
value  as  given  by  (6)  owing  to  the  back  e.m.f.  generated  by  the 
magnetic  field  of  the  inductance  coil.  For  a  short  interval  of 
time  after  closing  the  circuit  the  value  of  the  current  will  be 
given  by  the  expression: 

_Rt 
i  =  I  sin  (at  -  $  +  0)  -  Ie    L  sin  (^  -  0).  (7) 

In  short-circuiting  an  alternating  current  inductive  circuit,  in 
which  there  is  no  external  e.m.f.  acting  on  it,  the  current  grad- 
ually dies  out  in  accordance  with  the  expression: 

_Rt 
i=Ie    L  sin(^-0).  (8) 

Direct  Current  Circuits  Containing  Resistance  and  Capacity 
Only,  no  Inductance.  —  On  closing  the  circuit  the  current  and 
potential  difference  at  condenser  terminals  are  given  by  the  ex- 
pressions 

/-{•'»»,  o) 

i   ) 

(10) 

where  E  is  the  impressed  e.m.f.  and  R  and  C  are  the  resistance  and 
capacity  of  the  circuit.  Equation  (9)  does  not  apply  for  very 

E 

small  values  of  t,  because  at  t  =  0,  equation  (9)  gives  7  =  ^,  that 

is,  at  the  moment  of  closing  the  current  jumped  instantly  from 
zero  to  finite  value  and  this  is  not  possible.  To  get  the  value  of 
the  current  for  the  very  small  values  of  time  the  inductance  of 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    175 


the  circuit  must  be  taken  into  account  though  it  may  be  ex- 
tremely small,  and  in  that  case  we  get 

t  77F  R 

—  -57;         £j     —  •?  * 


Example. 


I  = 


R 


R 


E=  1000  volts, 
R  =  100  ohms, 
C  =  10  mf., 
L  =  1  mh. 


(ID 


t 

-RC 

R  t 

€      L 

e 

/ 

0 

1 

1 

0 

0 

io-4 

0.990 

0.368 

10 

6.22 

1(H 

0.904 

0 

96 

9.04 

2X10-4 

0.818 

.... 

282 

8.18 

5X10-4 

0.606 

394 

6.06 

8X10-4 

0.449 

551 

4.49 

10-3 

0.368 

632 

3.68 

1.5X19-3 

0.223 

777 

2.23 

2X10-3 

0.135 

865 

1.35 

3X10-3 

0.050 

950 

0.50 

4X10-3 

0.018 

982 

0.18 

5X10-3 

0.007 

993 

0.07 

It  is  to  be  noticed  that  the  factor  e  L  in  equation  (11)  is  only 
operative  for  an  extremely  small  value  time  interval,  and  after 
that  it  is  entirely  negligible  and  equation  (9)  may  be  used. 


CIRCUITS  CONTAINING  RESISTANCE,  INDUCTANCE  AND  CAPACITY 

IN  SERIES* 

The  problem  of  determining  the  current  and  potential  in  a 
circuit  containing  resistance  R,  inductance  L  and  capacity  C, 
either  on  charging  the  condenser  when  the  circuit  is  closed,  or  on 

*  W.  Thomson,  Transient  Electric  Currents,  Phil  Mag.,  Ser.  4,  Vol.  5, 
p.  393,  1853. 

J.  A'.  Fleming,  "The  Principles  of  Electric  Wave  Telegraphy,"  Ch.  I. 

C.  P.  Steinmetz,  "Theory  and  Calculation  of  Transient  Electric  Phenom- 
ena," Sect.  I,  Ch.  V. 

H.  Armagnat,  "The  Theory,  Design  and  Construction  of  Induction  Coils." 
Translated  and  edited  by  O.  A.  Kenyon. 


176  FORMULAE  AND  TABLES  FOR  THE 

discharging  the  condenser    when  the  circuit  is  short-circuited, 
resolves  itself  into  the  study  of  the  differential  equation 
<M         dl      I  _dE 

Ldt*~*  Kdt  +c~~dt' 

j-pi 
For  constant  e.m.f  .  ---  =  0  and  the  above  equation  reduces  to 


In  working  out  the  solution  of  equation  (13)  three  different 
cases  may  arise  depending  on  the  relative  magnitudes  of  L,  C 
and  R.  They  are  as  follows: 

(I)     R  >  2  y  ^  (logarithmic  case). 
(II)     R=2\/^  (critical  case).  (13) 

(III)     R  <  2  y  -~  (trigonometric  case). 

We  shall  give  below  the  formulae  for  the  current  and  potential 
for  all  three  cases  either  on  charging  or  discharging. 

CASE  I.    R  >  2  y  ~  ,  LOGARITHMIC  CASE 

On  charging,  the  current  in  the  circuit  and  the  voltage  across 
condenser  at  any  instant  of  time  are  as  follows: 

R-s  R+S 


R-S 

2L      -  (R  -  S)  e     *L         '         (15) 


where  S  • 

When  the  condenser  is  discharging  we  have  for  the  current  and 
condenser  voltage  the  following  equations: 


R~8* 


(17) 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    177 


where  EQ  is  the  condenser  voltage  before  the  condenser  is  begin- 
ning to  discharge. 

The  expressions  for  the  currents  on  charge  and  discharge  of 
condenser  are  identical  in  form  but  of  opposite  signs,  that  is,  the 
currents  are  exactly  the  same  but  in  opposite  directions.  The 
voltage  across  condenser  increases  in  one  case  in  exactly  the  same 
way  as  it  decreases  in  the  other. 


1.001  — 
0  9  1 

x  — 

^w 

0.8  1 
0.7  I 

f 

\ 

s 

^^^^^ 

^-^ 

,^- 

^>  ^" 

* 

* 

^spj* 

x^ 

^ 

Current  and  Potential  on 
Charging  Circuit  containing 
luctance  Capacity  and  Resistance 

,R>  2  1^~(  Logarithmic  Oas«) 

0  5  1      / 

t 

>< 

^ 

In. 

^ 

r 

x 

^ 

fi  0  1    / 

/ 

/ 

"^ 

^ 

^ 

0  2J  / 

/ 

•^^ 

•*•  — 

x 

If 

X 

2   4   6  8 


10  12  14  16  18 
"  t  X 

FIG.  55. 


20  22  24  26  28  30  32  34  36  38  40 
Seconds 


An  examination  of  the  expressions  for  the  current  (14)  and  (16) 
shows  that  the  current  has  zero  values  for  the  time  i  =  0  and 
t  =  GO  ;  hence,  at  some  instant  of  time  the  current  has  a  maximum 
value.  That  is,  the  current  is  unidirectional,  gradually  increasing 
from  zero  to  its  maximum  value  and  then  decreasing  to  zero  again. 
The  time  at  which  the  current  is  at  its  maximum  value  is  given 
by  the  expression 

L'      |±|.  (18) 


Example.  —  Suppose  we  have  a  circuit  of  the  following  constants: 
L  =  2  mh.,     C  =  2  mf .,     R  =  100  ohms,     E  =  100  volts. 


8 


8  x  10"3 


c 


^=5750, 


2x10 

=  44,250. 


=  77. 


178  FORMULAE  AND  TABLES  FOR  THE 

The  time  in  which  the  current  will  rise  to  its  maximum  value 
is 

L '       R  +  S      2  X  10~3        177 
t  =  -olog,  ^ ~  =  — == — loge-^-  =  5.3  X  10~5  sec. 

o          it  —  o  /  /  Z6 

CASE  H.    H2  =  —• ,  CRITICAL  CASE 

The  current  and  voltage  across  condenser  on  charging  are 
.       R 


(20) 
The  current  and  condenser  voltage  on  discharging  are 


CASE  HI.  R  <  2  y  ~i  TRIGONOMETRIC  CASE 

The  current  and  voltage  across  condenser  terminals  on  charging 
are 

/=-e-«<sin/ft,  (22) 


e  =  E  jl  -  e-  ««  ^cos  ^  +  -  sin  jsAj  ,  (23) 


where 


, 


For  condenser  discharge  we  have  for  the  current  and  condenser 
voltage.  the  following  equations: 


(24) 
61  =  Eoe-  at  (cos  $t  +  |  sin  j8  A  •  (25) 

In  this  case  also  the  currents  have  the  same  numerical  values 
on  charge  and  discharge  but  are  of  opposite  signs,  and  for  the 
condenser  potentials  the  increase  in  one  case  is  exactly  equal  to 
the  decrease  in  the  other  case. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    179 

Equations  (22)  to  (25)  represent  damped  periodic  curves,  that 
is,  the  current  and  potential  are  oscillatory  and  the  amplitudes 
are  continuously  diminishing.  The  maximum  value  of  the  current 
decreases  in  geometric  progression  as  the  time  increases  in  arith- 
metic progression. 

The  frequency  of  oscillation  is 

rr~ 

(26) 


The  frequency  decreases  as  the  resistance  in  the  circuit  is  in- 
creased and   becomes   zero,   ceases  to  be  oscillatory,   wheny- 

R2 
=  -—j-2   (critical  case).     When  the  resistance  is  small  so  that  R2 

can  be  neglected  in  comparison  with  -~  we  have 

L> 

n  =  — 1-7=-  (27) 

2irVLC 

7-> 

The  quantity  ^rj-  denoted  by  a  is  called  the  damping  factor. 

Z  h 

If  the  resistance  is  small  -  is  small  compared  with  unity  and  equa- 
tions (24)  and  (25)  reduce  to 

/  =-#0C/3e-«<sin#,  (28) 

ei  =  #06-°*  cos /ft;  (29) 

or  we  may  put 


I  =  #0y  ^e-*' cos /ft.  (30) 

Ic 

The  quotient  of  the  current  to  the  condenser  voltage  is  y  j.     If 

the  circuit  is  entirely  free  from  resistance,  the  oscillations  persist 
indefinitely,  and  they  are  generally  designated  as  undamped  or  free 
oscillations. 

The  ratio  of  the  amplitudes  of  successive  half  waves  is 

T  =  T  =  T  =  etc'  =  e"^'  (31> 

12          13          1* 


180  FORMULAE  AND  TABLES  FOR  THE 

rp 

where  T  is  the  duration  of  a  complete  period.     The  term  -~- 
=     -  =  -r—  is  denoted  by  5  and  is  called  the  logarithmic  decre- 


ment of  the  oscillation  per  half  period. 
Hence, 

7l  _/2  _/3_  _          . 

T   —  T   —  j  —  etc.  —  e     , 

J2         -13         i\ 

and 

8=  log,  £  =  10^,  etc.  (32) 

^2  ^3 

Since  n  =  -  ==  ,  approximately,  the  expression  for  the  loga- 

2  7T    v  Li\j 

rithmic  decrement  may  be  put  in  the  form 

(33) 

As  an  illustration  we  may  take  the  following  constants: 

L  =  1  mh.,    C  =  0.001  mf.,  R  =  100  ohms,  E  =  1000  volts. 


1 

is  negligible  compared  with  -=-„  . 


The  ratio  of  two  successive  half  waves  is 


5  =  loge  1.164  =  0.152. 


The  small  difference  in  the  values  of  5  arises  on  account  of  carry- 
ing only  to  two  decimal  places  the  values  of  e~at  in  calculating 
the  values  of  7i,  72,  etc. 

If  we  introduce  a  hot-wire  ammeter  in  the  oscillating  circuit, 
the  current  indicated  is  the  effective  value  of  the  square  root 
of  the  mean  square  of  the  discharge  current.  Generally  the  time 
interval  between  two  successive  discharges  is  sufficiently  long  to 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    181 

allow  complete  discharge  of  the  condenser.  That  is,  the  current 
is  practically  zero  before  another  discharge  takes  place.  To  get 
the  effective  value  of  the  current  we  may  therefore  integrate 
equation  (24)  from  zero  to  infinity  and  divide  by  the  time  interval 
between  two  successive  discharges.  The  square  root  of  the  value 
thus  obtained  will  be  the  effective  value  of  the  discharge  current. 
If  we  denote  by  N  the  number  of  discharges  per  second,  we 
have 

(34) 

If  the  condenser  is  charged  N  times  a  second  to  a  potential  EQ 
the  power  supplied  is 

.         •       (35) 


We  can  also  get  an  expression  for  the  effective  value  of  the  current 
in  terms  of  the  first  maximum  value  of  the  oscillation,  as  follows: 

INI&        i  ~ 

'eff  "     /  ~8ri   X  l      /«\«  '  (36) 

where  8  is  the  logarithmic  decrement,  n  is  the  frequency  of  oscil- 
lations and  N  is  the  number  of  discharges.  For  oscillatory  currents 
of  high  frequency  such  as  are  generally  employed  in  wireless 

telegraphy  -  is  generally  a  small  quantity  not  greater  than  0.1, 

7T 

and  often  much  smaller  than  that,  hence  we  may  neglect  (-)   in 

\7T/ 

comparison  with  unity,  and  equation  (36)  reduces  to 

rt 

Example.  —  Let  us  take  the  same  data  as   in  the   illustration 

given  on  page  180. 

EQ  =  100  volts,  C  =  10~3  mf.,  L  =  1  mh.,  R  =  100  ohms,  II  =  0.92, 

n  =^,  d  =  0.157. 

Z  7T 

Assuming  N  =  1000  we  have  by  (34) 

=  0.071  amp. 


182  FORMULAE  AND  TABLES  FOR  THE 

By  (36)  we  have 

e5  =  1.17. 
,   8  _  103  (0.92)2  X  1.17  1  5.04  X  10~3 

•°= 


=  5.04  X  10~3,  approximately. 
Jeff  =  0.071  amp., 

which  is  in  agreement  with  the  value  obtained  by  formulae  (34). 
It  is  also  to  be  noted  that  the  term  (-)  is  very  small  compared 


with  unity,  and  formula  (37)  would  have  given  the  same  results. 

Alternating  Current  Circuits  Containing  Resistance,  Induc- 
tance and  Capacity  in  Series.  —  When  an  alternating  e.m.f.  is 
introduced  into  a  circuit  having  a  resistance  R,  inductance  L  and 
capacity  C  the  current  and  condenser  voltage  do  not  attain  their 
permanent  values  instantly.  It  requires  a  certain  time  interval, 
which  is  generally  very  small  in  all  practical  cases,  before  the 
current  and  voltage  across  condenser  have  built  up  to  their  maxi- 
mum value. 

The  current  and  condenser  voltage  may  be  expressed  as  the 
difference  of  two  quantities, 


, 


where  the  subscript  s  denotes  the  steady  or  permanent  values  and 
the  subscript  v  denotes  only  the  variable  values  of  the  current 
and  voltage  which  are  present  during  the  transition  period.  Iv 
and  Ev  diminish  in  accordance  with  an  exponential  law  with  re- 
spect to  time  and  consequently  disappear  very  rapidly.  Formulae 
for  I8  and  E8  have  been  given  in  Chapter  IV,  hence  we  shall  give 
here  only  formulae  for  Iv  and  Ev. 

As  in  the  previous  section  we  have  to  distinguish  here  also 
three  distinct  cases  depending  on  the  relative  magnitudes  of  the 
electrical  constants  L,  C  and  R. 

Let  us  consider  first  the  trigonometric  or  oscillatory  case 


«<2W^ 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    183 

We  shall  assume  that  at  the  instant  of  closing  the  circuit  the 
steady  values  of  the  voltage  and  current  are 

Ea  =  #max  sin  (coZ+  ^)  =  #max  sin  ^=0, 


where 


sin  (wt  +  ^  —  0)  =  7max  sin  (^  — 


For  the  limiting  conditions  given  by  (39)  we  have  the  following 
formulae  for  /„  and  Ev: 

/L._.  ) 

(40) 

Ds^-^sin^+^+^^y-^sin^  j, 

PC  ) 

(41) 

where 


Any  change  in  the  electrical  constants  of  the  circuit  will  also 
introduce  a  transient  term.  The  current  and  potential  cannot 
pass  over  instantly  from  one  permanent  value  to  another,  it  re- 
quires a  certain  time  interval  for  the  readjustment.  During  the 
transition  period  the  current  and  potential  can  also  be  expressed 
by  formulae  (38),  but  in  this  case  the  values  of  Iv  and  Ev  are  as 
follows : 


E  sin  ft  +  A7  y  g  sin  (/^  -  7)  > ,  (42) 

^  =€-«'?  T^l^  sin  tf«  -  T)  +  ^  sin  /3^  ,  (43) 


where  A#  =  (E2i, 

A/  =  (72>8  -  Ji,.)(«=o). 

Equations  (44)  give  the  variation  in  the  permanent  values  of 
the  current  and  condenser  potential  in  the  circuit  at  the  time 
t  =  0. 


184 


FORMULAE  AND  TABLES  FOR  THE 


For  the  critical  case  R  =  2  y  -^  the  value  of  Iv  is  given  by  the 
formula 


Iv  =  €- 


cos  *-*- 


(45) 


Closing  on  Alternating 
Inductive  Circuit 


FIG.  56. 


When  the  resistance  is  still  further  increased,  so  that  R  >  2  y  -^, 
we  have  the  logarithmic  case,  and  the  value  of  Iv  is 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    185 

[X 


<l/£«.m«COS   W  — 

^c.max  COS  (\I/  —  d>)-H  TT  +^j»  Umax  Sin  M  — < 


(46) 


As  in  the  trigonometric  case  we  have  here  also  for  any  change 
in  load  condition  a  transition  period  during  which  we  have,  be- 
sides the  term  representing  the  permanent  value  of  the  current, 
another  term  which  is  effective  for  a  short  period  only,  the  transient 
term.  The  value  of  the  transient  term  which  we  designated 
throughout  by  /„  is  as  follows: 


(47) 
^»(L  \*  /       J  L  \*  /       J          ) 

where 


-«  \ 


1 

J 


(48) 


represent  the  differences  in  7  and  #  between  the  steady  values 
before  and  after  the  change. 
The  values  of  a  and  /?  in  equations  (46)  and  (47)  are  given  by 


Divided  Circuits.*  —  We  shall  first  consider  the  simplest  case, 
the  circuits  having  only  inductance  and  resistance. 

Suppose  we  have  a  continuous  current  flowing  through  the 
circuits  and  either  the  main  circuit  is  short-circuited,  or  a  varia- 
tion in  the  resistance  of  any  of  the  branches  is  made.  The  problem 
is  to  determine  the  values  of  the  transient  terms  of  the  currents 
in  the  branches. 
Let  E  =  e.m.f. 

LO,  Li,  L2  =  inductances  of  main  and  branch 

circuits  respectively. 
RO,  Riy  R2  =  resistances  of  main  and  branch 

circuits  respectively. 
The  permanent  values  of  the  currents  in  branches  1  and  2  are 


T  ,     r  t    N 

~W  2  =  ~W'  ^   ) 

where  R*  =  RoR2  +  R^ 


*  See  Ch.  P.  Steinmetz,  "Theory  and  Calculation  of  Transient  Electric 
Phenomena  and  Oscillations,"  Section  I,  Ch.  IX. 


186 


FORMULAE  AND  TABLES  FOR  THE 


Now  let  us  suppose  that  there  is  a  change  in  resistance  in 
branch  1  or  2.     The  currents  in  the  two  branches  are  as  follows: 


(50) 


T  I" 


X. 


... 


(51) 


where  //  and  72'  are  the  permanent  values  of  the  currents  in  the 
two  branches  before  the  change  in  the  resistance,  and  //'  and  /2" 
are  the  permanent  values  of  the  currents  after  the  change. 


Rv 


00000 


FIG.  57. 


=  L0L2 


4"  RoRi  H~ 


2L2 


(L,  +  Lo), 


2L2 


As  an  illustration  let  us  consider  the  following  example: 

LO  =  5  henrys,      LI  =  2.5  henrys,        L2  =  10  henrys, 
RO  =  10  ohms,       RI  =  2  ohms,  Rz  =  25  ohms, 


and  we  shall  assume  that  the  value  of  Rz  is  suddenly  changed  to 
15  ohms. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    187 

The  permanent  values  of  the  currents  before  and  after  the 
change  in  the  resistance  are 

//  =  7.81  amp.,  h'  =  0.625  amp., 

/i"  =  7.5  amp.,  /,"  =  1.00  amp. 

We  also  have 

L2  =  87.5,  jR2  =  200,  /S2  =  267.5. 

Xi=-  1.42,  X2=-1.63, 

Ki  =  0.47,  K2  =  -  0.12, 

Introducing  these  values  in  equations  (50)  and  (51)  we  get 
Ji  =  7.5  +  0.70  e-1-42'  -  0.39  e-1-63', 
72  =  1.00  -  0.33  e-1-42'  -  0.045  e-1-63'. 

For  t  =  0 

which  are  the  permanent  values  of  the  current 
before  the  change  in  resistance. 

For  t  =  oo 

71  =  7.5    amp.  "1      which  are  the  permanent  values  of  the  current 

72  =  1.00  amp.  J      after  the  change  in  resistance. 

If  the  circuit  is  short-circuited  we  must  put  in  equations  (50) 
and  (51)  /i"  and  72"  separately  equal  to  zero,  and  the  equations 
reduce  to 

,       1^2/1'  +  Izf  x  ,      Kill  +  ^2    XJ 
fi  =  ~  -  Y~  e     --  -    ~c    » 


/!  =  7.81     1 
72  =  0.625   J 


7     -       _        lzl  l%      *t     ,          l^z    H"  1^2/2^    x  j 

*2  JV-  7^  ^  T  T7-  T7-  ^        ' 

Xvi  —  /V2  -tVi  —  /V2 

Using  the  same  constants  as  in  the  above  example,  we  get 

/i  =  0.53  c-1-42^  7.28  C-1-63', 
72  =  -  0.25  e-1-42'  +  0.87  e-1-63'. 

Transient  Currents  in  Transmission  Lines.*  —  The  capacity 
of  the  transmission  line  is  represented  by  a  condenser  shunted 
across  the  middle  of  the  line.  We  shall  give  here  formulae  for 
the  transient  currents  and  potentials  which  may  arise  either  when 
the  load  is  short-circuited  or  the  generator  is  short-circuited. 

*  See  J.  L.  LaCour  and  O.  S.  Bragstad,  "Theorie  der  Wechselstrome," 
Ch.  XXIV. 


188 


FORMULAE  AND  TABLES  FOR  THE 


Consider  first  the  case  of  the  transient  phenomena  which  occur 
in  transmission  line  when  the  load  is  short-circuited,  as  shown 
diagrammatically  in  Fig.  58. 


FIG.  58. 

Let  71>8)0,  7a,.,0,  7C,S>0,  ECt8>0  represent  the  permanent  values  of 
the  currents  in  the  different  branches  and  the  condenser  potential 
at  the  instant  that  the  line  is  short-circuited  at  the  receiving  end. 
Then  we  have  the  following  formulae  for  the  transient  values  of 
the  currents  and  potential. 
W 

(54) 


(55) 


=  € 


(56) 


where 


a  = 


Ei        Rz 
2Li"2L2 


8=,i  =    -2  =2o:      and     0  =  tan-1- 
Li      Lz 

The  values  of  a  and  /3  given  above  have  been  obtained  on  the 

. .       , ,     ,  RI      Rz 
assumption  that  -j—  =  j — 
LI      L2 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    189 

An  examination  of  equation  (56)  makes  it  appear  evident  that 
no  great  rise  in  potential  can  occur  on  the  line. 

As  an  illustration  consider  the  case  of  a  single-phase  line  100 
miles  longx  made  up  of  No.  000  B.  &  S.  with  wires  96  inches 
interaxial  distance.  The  resistance  per  mile  R  =  0.33  ohm. 

inductance  per  mile  L  =  2.06  mh. 
Capacity  per  mile  C  =  0.0145  mf . 

The  total  resistance,  inductance  and  capacity  of  the  line  are 
33  ohms,  206  mh.  and  1.45  mf.  respectively. 
We  have  therefore 


Lj  =  L2  =  103  mh.      a  =  -—-  =  80.1, 

'  •  V/103  X  10-3  X  1.45  X  1**  ~  (8°'1)2  "  36'6  X  10 ' 
/3C  =  0.005, 


=      l.004  =  1,  approx. 


EC.,  =  -  c-80  «  |  sin  ft  +  EC,,,Q  sin  (ft  + 

=  -  c-80  «  1  200  7C,8>0  sin  ft  +  Ec,8>0  sin  (pt  +  4>)  I  . 
If  7c>8,o  is  numerically  equal  to  one  hundredth  of  Ec>8i0,  that  is, 

ET 

7Ci8>0  =  -jg^j  the  rise  in  potential  is  only  about  three  times  the 

normal  value. 

For  the  case  where  the  generator  is  short-circuited  we  may 
neglect  L2  compared  with  R2,  and  in  that  case  the  transient  cur- 
rent in  the  branches  and  the  potential  is  given  by  the  following 
expressions: 


*)|j  (58) 
1  sin $  +  £e,.,o  V  1  +  I* sin  (|M  +  0)  K        (59) 


190 
where 


FORMULAE  AND  TABLES  FOR  THE 


a 


~h 


tan 


_  ,v2 


FIG.  59. 


If  a  direct  e.m.f .  is  acting  on  the  line,  then  ECta>Q  =  E  and  7c>8i0  =  0, 
where  E  is  the  e.m.f.  impressed  on  the  line.  Equation  (59) 
reduces  to 


=  6 


y  1  +      sin 


E 


The  increase  in  potential  is  proportional  to 


(60) 


This  ratio  is  seldom  much  greater  than  unity  and  increases  as  a 
increases. 

Mutual  Inductance.*  —  If  we  have  two  circuits  connected 
inductively  and  impress  continuous  e.m.f.  Js  E\  and  E2  on  circuits 
I  and  II  respectively  the  currents  in  the  circuits  are 


and 


(61) 


*  See  Ch.  P.  Steinmetz,  "Theory  and  Calculation  of  Transient  Electric 
Phenomena,"  Section  I,  Ch.  X. 

Einfiihrung  in  die  Maxwellsch  Theorie  der  Elektrizitat  von  Dr.  A.  Foppl 
herausgegeben  von  Dr.  M.  Abraham,  Part  III,  Ch.  II. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    191 


If  there  is  a  sudden  change  in  load  conditions,  say  R%  is  changed  to 
Rz',  the  permanent  values  of  the  currents  in  the  circuits  after  the 
change  are 


(62) 


Ii"  =  g     and 

The  change,  however,  from  one 
current  value  to  another  due  to 
a  change  in  circuit  conditions 
does'not  occur  instantly.  The  cur- 
rents rise  gradually  to  the  changed 
permanent  values;  the  interval 
of  time  required  before  the  perma- 
nent conditions  are  established 
again  depends  upon  the  electric  constants  of  the  circuits. 

The  expression  for  the  currents  in  the  two  circuits  at  any  time 
t  are  as  follows: 


(63) 


FIG.  60. 


where 


JLi  —  J-i    -f  -           —  e"1* 
ra<3  —  mi 

„        miMz    ,t  , 

m2  —  mi 
m2A72    x 

±2          1-2                                     €  l    - 

W2  —  mi 

T    --  T  f         T  ff 
i  l2  —  1%    —  J-2    , 

m2  —  mi      ' 

m\i 

mX2      ' 

^             -    («1  +  0!2)   +  V(a 

2  -  <*i)2-f  4oW£2 

1- 

X2 

2  -o:i)2H-4Q:1a2X2 

1- 

7?                               J? 
-ftl                            .fi/2 

«i  =  2j-  ,       a2  =  2^-? 
IT.  -   M2  . 

X2 

(64) 


(65) 


If  there  is  a  change  in  the  resistance  of  circuit  I  from  RI  to 
Ri  the  permanent  values  of  the  currents  in  the  circuits  after  the 
change  are 


and 


(66) 


192  FORMULAE  AND  TABLES  FOR  THE 

The  currents  in  the  circuits  before  the  permanent  values  are 
established  are  given  by  the  expressions. 


_        _  _  eXl(  +  __  £M  (67) 

mi  —  m?  mi  —  mz 


=  A7.rn.m3  fXil  _  ^ 

mi  —  m2  mi  —  mi 

where  A/i  =  Ii  —  Ii"  and  mi,  m2,  Xi,  \2  have  the  same  signifi- 

cance as  given  in  equation  (65). 

If  one  of  the  circuits  is  suddenly  opened,  we  shall  merely  have 

to  put  either  72"  =  0  in  equations  (63)  and  (64)  or  7/'  =  0  in 

equation  (67)  and  (68)  depending  upon  whether  we  open  circuit  II 

or  circuit  I. 

Example.  — 

Li  =  0.1  henry,  Ri  =  10  ohms,  L2  =  0.2  henry, 
#2  =  25  ohms,  M  =  0.1  henry,  EI  =  100  volts, 
E2  =  200  volts. 

Now  suppose  we  suddenly  change  R2  to  40  ohms.     We  have 

100  r  ,      200  7  „      200     K 

hf  =  -jQ-  =  10  amp.,      U  =  -^  =8  amp.,      72"  =  -^  =5  amp., 

10  25 


-  (50  +  62.5)  +  V(62.5  -  50)2  +  4X  50X  62.5 X  0.5  _ 
=~  1-0.5  " 


_  (50  +  62.5)  -  V (62.5 -50)2+4X5QX62.5X0.5^ 

1  -0.5 

_  -  0.1  X  65  +  10       _  -  0.1  X  385  +  10  = 

-  0.1  X  65  -  0.1  X  385 

A72  =  8-5  =  3; 

hence,  7i  =  10  +  ^  e~*5 '  —  j^g  €~385 '  > 

72  =  5  +  1-26  e-65'  +  1.74€-385'. 
For  t  =  0 

71  =  10  amp.  \  permanent  values  before  change  in 

72  =  8  amp.    J      resistance  of  circuit  II. 

For  t  =  oo 

7i  =  10  amp.  \  permanent  values  after  change  in 
Iz  =  5  amp.  J     resistance  of  circuit  II. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    193 

If  in  the  above  example  we  had  changed  R\  from  10  to  20  ohms 
and  kept  R2  constant,  we  should  have  had 


7,'=         =  10  amp.,     /1"  =          = 

mi,  w2,  Xi,  X2  have  the  same  values  as  above. 
By  (67)  and  (68)  we  get 


72  =  8  +  1.56e-*5<-  1.56  e-385'. 
For  t  =  0 

7/  =  10.  amp.  1  permanent  values  before  the  change  in 
1  2  =  8    amp.  J       resistance  of  circuit  I. 
For  t  =  oo 

7i"  =  5  amp.  j  permanent  values  after  the  change  in 
72"  =  8  amp.  J      resistance  of  circuit  I. 

Let  us  now  consider  the  case  where  one  circuit  has  no  e.m.f. 
acting  on  it  and  an  e.m.f.  EI  is  introduced  in  the  other  circuit. 
Suppose  E%  =  0.  There  is  no  current  in  either  circuit  before 
the  e.m.f.  is  introduced  in  cifcuit  I;  that  is,  we  have 

7/=0,     72'=0,     Ii"=^>    72"  =  0. 

Jtti 

Equations  (67)  and  (68)  reduce  for  this  case  to  the  following: 

"  (69) 


The  values  of  mi,  m^  and  Xi,  X2  are  given  by  (65).  The  current 
in  the  secondary  circuit  first  commences  to  rise,  reaches  a  maxi- 
mum value  and  then  diminishes  to  zero  again.  The  time  in 
which  it  reaches  its  maximum  value  is  given  by 

«  =  —  Lj-Io6^-  (71) 

X2  —  Xi         X2 

As  an  illustration  we  may  use  the  same  constants  as  in  the  pre- 

vious example, 

•p 

mi  =-0.54,    m2  =  0.74,     Xi  =  -65,     X2=-385, 


7i  =|10  -  5.78€-*5<-  4.22<r385'S,    h=-  3.12 


194  FORMULAE  AND  TABLES  FOR  THE 

The  time  in  which  the  secondary  will  have  its  maximum  value  is 
t  =  ^  X  1.78  =  0.0056  sec. 


The  table  below  gives  the  values  of  /i  and  I*  for  values  of  t 
(0  -  0.015)  calculated  by  formulae  (69)  and  (70). 


t. 

/i. 

/I. 

0 

0 

0 

0.001  sec. 

1.72 

0.80 

0.002 

2.97 

1.29 

0.005 

5.21 

1.80 

0.008 

6.36 

1.71 

0.010 

6.9 

1.56 

0.015 

7.81 

1.16 

Oscillation  Transformer  Having  Inductance  and  Capacity  in 
Series.*  —  The  complete  solution  of  the  problem  of  determining 
the  current  and  condenser  potential  in  each  circuit  of  the  oscilla- 
tion transformer,  when  an  electric  discharge  occurs  in  one  circuit, 
offers  considerable  mathematical  difficulties.  It  involves  the  solu- 
tion of  a  fourth-degree  differential  equation  of  the  form 


(LJ*  -  M2)        +  CiC,  (WZ,  + 

)       +  ( 


+7  =  0      (72) 


where  Ri,Li,Ci  and  ^2,L2,C2  are  the  resistances,  inductances  and 
capacities  of  the  primary  and  secondary  circuits  respectively  and 
M  is  the  mutual  inductance  between  the  two  circuits.  Since, 
however,  in  most  cases  where  such  circuits  are  employed,  as  in 
the  case  of  a  radiotelegraphy,  the  resistances  are  small  com- 

*  J.  A.  Fleming,  "The  Principles  of  Electric  Wave  Telegraphy  and 
Telephony,"  Ch.  III. 

P.  Drude,  "tiber  induktiv  Erregung  zweier  elektrische  Schwingungskreise, 
etc.,"  Ann.  der  Physik,  1904,  Vol.  13,  p.  512. 

A.  Oberbleck,  "tiber  den  verlauf  der  elektrischen  Schwingungen  bei  den 
Tesla'schen  Versuchen,"  Wied,  Ann.  der  Physik,  1895,  Vol.  55,  p.  623. 

Louis  Cohen,  "Theory  of  Coupled  Circuits  having  Distributed  Inductance 
and  Capacity,"  Bulletin  Bureau  of  Standards,  Vol.  5. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    195 


pared  with  the  reactances,  we  may  as  an  approximation  neglect 
Ri  and  R2  which  reduces  equation  (72)  to  a  second-degree  equa- 
tion. With  this  limitation,  that  is,  neglecting  the  resistances,  we 
have  for  the  condenser  voltages  of  the  primary  and  secondary 
circuits 

C, 


FIG.  61. 


a2 


1 02  COS  U\t  —  Oi  COS  Uzt  I , 


where 


1  - 


02  = 


(73) 


(74) 


EI  is  the  maximum  voltage  of  condenser  Ci,  that  is,  the  break- 
down voltage  of  the  gap,  and  coi,  co2  are  2  TT  times  the  frequencies 
of  the  circuits.  Both  circuits  oscillate  with  two  periods  corre- 
sponding to  the  frequency  constants  coi  and  co2. 

The  values  of  «i  and  coi  are 


=  4  / 


-  C2L2)2 


2CiC2(L1L2-M2) 


W2  = 


2  CiC2  (LiL2  -  M2) 
When  the  two  circuits  are  syntonized 


Equations  (75)  reduce  to 


(75) 


V  LC(l- 


(76) 


where  K  =  —f and  is  called  the  coefficient  of  coupling. 


196  FORMULAE  AND  TABLES  FOR  THE 

If  we  put  CL  =  -5  where  —  is  the  natural  time  period  of  each 

CO2  CO 

circuit,  we  have 

CO  CO 

= 


Substituting  the  values  of  ai  and  (h  from  (74)  into  the  second 
equation  of  (73)  we  get 


_  _  _  LiI/2  __          ,__, 

"  02  -  ai      V(LA  -  L2C2)2  * 


When  the  two  circuits  are  syntonized,  we  get 


Ivi- :  (78) 


2  VC2 
If  ^  =  1  or  M2  =  LiI/2  equation  (77)  becomes 


The  current  in  the  secondary  circuit  is  also  expressed  as  the 
sum  of  two  currents  of  different  frequencies. 


(80) 
The   oscillation   of  the   greatest   frequency   has   the  greatest 

amplitude. 
As  an  illustration,  we  take  the  following  constants: 

Li  =  2  mh.,  L2  =  l  mh.,  Ci  =  0.002  mf  .,  C2  =  0.001  mf  .,  M  =  1  mh. 

We  have 

Lxd  =  4  X  10-12,    L2<72  =  10-12,    4  M2CA  =  8  X  10~24, 
CA  (ZaZ*  -  M2)  =  2  X  10-18  (2  X  lO-6  -  1(H)  =  2  X  10~24, 

hence 


^______^_ 

»L    JLt/5  x  io-12  +  v^xTo-84  +  8  x  i(F 

V 


4  X  10~24 
==  2.41  X  IO5, 


s-  VT7  =  0.74  X  IO5. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    197 

The  two  frequencies  are  in  the  ratio  of  3.26  :  1. 
If  the  two  circuits  are  syntonized,  that  is,  if 

L2C2  =  LA  =  4  X  10-12    and    £  =  0.5, 
we  have 

W2         1   t/ 1 1Q6        -  9 

711  =  2^  =  2i  V  lO-i*  1-0.5    ~  o 


712 


"2  Li/I         _J 

"  2^  "  2  TT  V  10-12  (1  +  0. 


(1+0.5)      2irVO 
and  the  current  in  the  secondary  circuit  is  as  follows; 

(n  \s  -i  f\ — 9  O  y  -I  f\ — 9 

/                              ^  \/  O  OK  V  1  (\!a  ain  /•     /  *^ \S  1    Q  \ 
^ /\  ^i.^O  /\  1U    bill  COll/  ~~             pC  /x  -L«<J  / 
Z                                                                                         ^ 

=  Ei  X  10~4  (2.25  sin  out  -  1.3  sin  «iO- 

The  wave  lengths  can  be  obtained  from  the  values  of  the  fre- 

y 
quencies  by  the  relation  X  =  — ,  where  V  is  the  velocity  of  light; 

Y  =3X1010  —  • 
sec. 

For  the  case  in  the  above  example  when  the  two  circuits  are 
syntonized  we  have 


q  y  -j  mo          q 

X2  =  =        X  105  cm.  =  2307  meters. 


Direct  Coupling.  —  Another  form  of  coupling  commonly  used 
in  high  frequency  work  for  the  generation  of  oscillatory  currents 
is  the  so-called  direct  coupling.  The  arrangement  of  the  circuits 
is  shown  in  Fig.  62.  When  an  electric  discharge  occurs  in  cir- 
cuit I,  the  current  is  partly  transmitted  to  circuit  II  which  in 
its  turn  reacts  on  circuit  I  and  thus  both  circuits  are  set  in 

*  G.  Seibt,  Physikdische  Zdtschrift,  Aug.  1,  1904;  or  Ed.  Elect.,  Oct.  1, 
1904. 


198 


FORMULAE  AND  TABLES  FOR  THE 


electrical  oscillation  simultaneously.  As  in  the  case  of  the 
magnetically-coupled  circuits,  both  circuits  oscillate  with  two 
different  frequencies. 

C, 


FIG.  62. 

If  we  arrange  the  constants  of  the  circuits  so  that  both  are  in 
syntony  we  have 

Then  the  frequencies  of  the  oscillations  are 


n2  =  x~ 


(82) 


(83) 


If  we  write  in  the  above  equation 

equations  (82)  and  (83)  reduce  to 

1 


1  — 


instead  of  1  +  j^ » 

(84) 

2  TT  VLC  VI -p 
If  jl  is  small,  p  is  also  small  and  the  difference  in  the  two  fre- 
quencies will  be  also  small.     As  an  illustration  we  may  consider 
the  case  of  antenna  circuit  coupled  directly  to  the  exciting  circuit 
as  shown  in  Fig.  63. 

We  may  use  the  following  constants: 

Li  =  0.1  mh.,    L2  =  1  mh.,     d  =  0.004  mf., 


=  0.1,     1  -  p2  =  0.9,    p2  =  0.1  and  p  =  0.316. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    199 

Introducing  these  values  in  equations  (84)  and  (85)  we  get  . 

1  1 


2  TT  V  10~4  X  4  X  10-9   V1.316 

1 1 

27T VlO-4X4  X  10-9  Vo.684 


=  2.2  X  105, 
=  3.05  X  105. 


FIG.  63. 

The  wave  lengths  corresponding  to  these  frequencies  are 
Xi  =  1364  meters. 
X2  =  984  meters. 

If  in  the  above  example  we  had  L2  =  2  mh., 


=  0.05,    j-i—  2-1.05,     P2  =  0.05,     P  =  0.224, 


and 


27rV10-4X4X10-9          T224 


2  TT  V  10~4  X  4  X  10-9         0.776 


=  2.27  X  105, 


=  2.86  X  105. 


200  FORMULAE  AND  TABLES  FOR  THE 


CHAPTER  VI 
DISTRIBUTED    INDUCTANCE   AND    CAPACITY* 

IN  long-distance  transmission  problems  it  is  not  permissible  to 
assume  that  the  resistance,  inductance  and  capacity  are  localized  at 
one  or  more  points.  The  error  introduced  by  such  an  assumption 
may  be  considerable,  as  will  be  shown  by  some  examples  worked 
out  here.  Accurate  and  reliable  results  can  be  obtained  only  by 
working  out  the  problem  in  its  most  general  form,  allowing  for  the 
uniform  distribution  of  resistance,  inductance,  capacity  and 
leakage  along  the  entire  length  of  the  line.  The  problem  does 
not  offer  any  great  analytical  difficulties;  it  has  been  discussed  by 
many  engineers  and  physicists  and  various  formulae  have  been 
derived.  The  numerical  work,  however,  in  working  out  any  par- 
ticular problem  by  the  general  formula  is  somewhat  laborious. 
It  is  hoped  that  the  various  tables  and  curves  worked  out  here 
will  facilitate  considerably  the  work  of  numerical  computation. 

More  than  one  formula  are  given  in  many  cases  so  that,  by 
carrying  through  the  computations  by  two  different  formulae, 
numerical  errors  may  be  eliminated. 

It  is  believed  that  the  many  examples  covering  typical  cases 
of  transmission  lines  worked  out  here  will  prove  to  be  of  value 
to  engineers  who  have  to  deal  with  transmission  problems,  in 
giving  an  approximate  idea  of  the  general  behavior  of  a  long- 
distance transmission  line  in  regard  to  its  electrical  conditions  as 
to  losses,  regulations,  etc.  Furthermore,  some  examples  have 
also  been  worked  out  by  other  formulae  which  have  been  de- 
rived on  the  assumption  of  localized  inductance  and  capacity, 
which  will  serve  to  illustrate  what  errors  may  be  expected  in  the 
use  of  such  formulae. 

Notation.  —  In  all  the  formulae  in  this  chapter  we  shall  use  the 
following  notation: 

R  =  resistance  in  ohms  per  unit  of  length  (the  mile  is  a  con- 
venient unit  of  length), 
*  A  list  of  references  on  this  subject  is  given  at  the  end  of  the  chapter. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    201 

L  —  inductance  in  mh.  per  unit  of  length, 
C  =  capacity  in  mf .  per  unit  of  length, 
g  =  leakage  conductance  in  mho.  per  unit  of  length, 
co  =  2  TT  X  frequency, 
n  =  frequency, 

s  =  distance  along  the  line  either  from  generator  end  or  re- 
ceiving end, 

Eg  =  voltage  at  generator  end, 
Ig  =  current  at  generator  end, 
Er  =  voltage  at  receiving  end, 
I  r  =  current  at  receiving  end, 
E  =  voltage  at  any  point  on  the  line, 
7  =  current  at  any  point  on  the  line, 
a  =  attenuation  constant, 
j8  =  velocity  constant. 


The  General  Equation  for  Current  Propagation  Along  Wires. 

• —  The  general  differential  equations  for  the  current  and  poten- 
tial distribution  along  a  line  of  uniformly  distributed  inductance 
and  capacity  are  as  follows: 


i+v-E, 


\dt2 


ds2 


(1) 


Since  we  are  to  consider  only  sinusoidal  quantities  (7 
E  =  Eot3'"*),  these  equations  may  be  simplified  to 


ds* 


where 


(2) 


F2  =  -  LCco2  +  (CR  +  Lg)  ju  +  Rg.  (3) 

The  interconnections  between  E  and  I  are  given  by  the  equations : 


(4) 


202  FORMULAE  AND  TABLES  FOR  THE 

The  plus  sign  is  used  in  considering  distance  from  receiving  end, 
that  is,  in  direction  of  increasing  power.  The  minus  sign  is  used 
in  counting  distance  from  generator  end,  that  is,  in  the  direction 
of  decreasing  power. 

In  equations  (2)  V  is  the  propagation  constant  of  the  line.    It 
is  a  complex  quantity  and  may  be  put  in  the  form 


where 

a  =  Vj  {  V(S2  +  co2L2)  (g*  +  CVO  +  Rg  -  co2LC)  }  , 
/  -  .  - 

0  =  V  J  }  V(R*  +  «2L2)  (g*  +  C2co2)  -Rg  +  co2LC)  | 


a  is  the  attenuation  constant  giving  the  rate  at  which  the  current 
and  potential  waves  diminish  in  intensity  as  they  travel  along  the 
line,  and  is  a  function  of  the  frequency.  The  higher  the  fre- 
quency the  more  rapid  the  damping.  This  is  very  objectionable 
in  telephonic  transmission  since  the  higher  harmonics  of  the  speech 
wave  are  diminished  in  intensity  at  a  more  rapid  rate  than  the 
lower  harmonics,  and  consequently  the  wave  arriving  at  the  re- 
ceiving end  is  distorted. 

j8  is  the  wave  length  or  velocity  constant  of  the  line.  If  we 
designate  by  X  the  wave  length  and  by  W  the  velocity  of  prop- 
agation we  have 

TF=?p,    X=|-  '  (6) 

Example.  — 

Consider  a  telephone  circuit,  metallic  return,  consisting  of  No. 
14  B.  &  S.  copper  wire,  interaxial  distance  24  ins. 

We  have 

R  =  13.3  ohms  per  mile, 

L  =  2.21  mh.  per  mile, 

C  =  0.0135  mf  .  per  mile. 
Assume      g  =  0,  leakage  negligible. 
For  co  =  1000,  R2  +  co2L2  =  181.77,     Ceo  =  13.5  X  10"6, 


X  10-«  (V181.77  -  2.21)  =  8.72  X  10~3, 


=  X  10-«  (  VI8L77  +  2.21)  =  10.29  X  10~3. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    203 
The  velocity  of  propagation  W  =  ^  =  1  n  Q       n  n_3  =97,100  miles 

p        JLU.o  /x  -i-U 

TF      TF  X  2  TT 

per  second.     The  wave  length  X  =  —  =  -  -  -  =  609.7  miles. 

n  /3 

If  we  take 
co  =  10,000,        R*  +  co2L2  =  665.3,        Ceo  =  135  X  10~\ 


x  10-*(6653  -  22.1)  =  15.78  X  10~3, 


x  iQHi(665     +  22.1)  =  56.85  X  10~3, 
W  =     =          10 


Increasing  the  frequency  ten  times  has  about  doubled  the  value 
of  the  attenuation  constant,  and  decreased  the  wave  length. 

If  we  neglect  the  leakage  and  make  L  large  by  the  introduction 
of  loading  coils  at  various  points  on  the  line,  then  we  may  write 
equations  (5)  in  the  following  form  : 


y ,  approx. 


(7) 


The  attenuation  is  practically  independent  of  the  frequency. 
If  the  constants  of  the  line  are  of  such  magnitude  as  to  make 

J?.  =  JL 

Leo       Ceo 
equations  (5)  will  reduce  to 


a  = 


|     and     0  =  «y  CZl  +  -j^-2  =  co  VCL,  approx.  (8) 


In  this  case  the  attenuation  is  also  independent  of  the  frequency, 
but  it  is  to  be  noticed  that  in  (8)  the  attenuation  constant  is 
twice  as  large  as  that  in  (7). 

For  telephonic  cables  when  the  wires  are  close  together,  the  in- 
ductance is  negligibly  small,  and  equations  (5)  reduce  to 


204  FORMULAE  AND  TABLES  FOR  THE 


(9) 


When  the  leakage  component  is  small  compared  with  the  capacity 
component 


This  is  the  formula  which  ordinarily  applies  to  cable  circuits 
employed  for  telephonic  transmission. 

We  may  also   express  a  and  /3  in  the  following  form : 


co  X  3.785 

a  — 


(11) 

/-.i  V  2  785       /   /  1        \    /     /  no  \ 

0 

where  D  =  interaxial  distance  between  the  to  and  fro  conductor 

and  r  is  the  radius  of  the  conductor.     Since  L  is  a  function  of  — , 

r 

we  may  express  a  and  /3,  for  a  given  resistance  and  a  given  fre- 
quency, as  a  function  of  — 

Formulae  (11)  were  used  for  the  calculation  of  Tables  XVII 
and  XVIII  and  Figs.  64-67  were  plotted  from  these  tables, 
giving  the  values  of  a  and  0,  for  the  frequencies  25  and  60  cycles, 

as  functions  of  — 
r 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    205 


I  E 

^  « 

9  5 


I 


llSIIISSiil 

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co  c<  c  e<  c< 


Mw 


Illllllliil 

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o  o  o  o  o  o 


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SOeoeo«o«ooc^w>oo» 
COOOCOCO*O^fCOC^i—  1O 


?  i  i  j  3,  3, 


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C^ 


t^t^^HC5 


i-  ^-  0  0  0  0  0  c>  0  G>  0 


XXXXXXXXXXX 


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i-IOOOOOOOOOO 


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TtftlTTTTTT 

S22SSSSS2S3 

d  o  d  o  o  d  d  o  d  d  d 


T-HC3COIOOOOO 


206 


FORMULAE  AND  TABLES  FOR  THE 


a  a 
>  S 


S 


«  3 

S  § 

B 

e 


13 


IS 


2| 


wa 

8-S 


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IIIIIIIIISI 

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T-l  i-l  T-l  rH  ^H  T-I  d  O  O  O  O 


x  3, 


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0  0  0  O  O  O  O 


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o  o  o  o  o  o  o  o  o  o  o' 


xx 

cqio 


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cot>-ooOi>-OiO5cooo 


kl 
XX 


Illll 

X  X  X  X  X 


illl 

X  X  X  X 


o  o  o  o  o  o  o   o  o  o 


xx 


IIIIS 

X  X  X  X  X 


llll 

XX  XX 


^Hr-t.-iOOOO 


CS  O>  OS  O> 

d  d  d  o 


II  „, 

X  X  X  X  X  X  X 


lll 


Illl 

X  X  X  X 


o  o  o  o  o   o  o  o   o 


aa88 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    207 


TABLE  XIX* 

VALUES  OP  a  AND  /?  PER  MILE  FOR  TRIANGULAR  SPACING  OF  THREE-PHASE 
TRANSMISSION  LINES  AT  25  AND  60  CYCLES 


Size  of 
wire, 
B.  &S. 

Resist- 
ance in 
ohms 
per  mile 

Spacing 
between 
wires  in 
inches. 

60  Cycles. 

25  Cycles. 

a 

0 

a 

0 

No. 

250,000 

0.222 

(  72 
j    96 
1  120 
1.144 

0.322X10-3 
0.306X10-3 
0.294X10-3 
0.287X10-3 

2.11X10-3 

2.10X10-3 
2.09X10-3 
2.09X10-3 

0.306X10-3 
0.306X10-3 
0.283X10-3 
0.275X10-3 

0.918X10-3 
0.912X10-3 
0.907X10-3 
0.906X10-3 

0000 

0.263 

(  72 
)    96 
1  120 
1144 

0.374X10-3 
0.357X10-3 
0.344X10-3 
0.334X10-3 

2.11X10-3 
2.10X10-3 
2.10X10-3 
2.09X10-3 

0.352X10-3 
0.337X10-3 
0.326X10-3 
0.316X10-3 

0.932X10-3 
0.927X10-3 
0.922X10-3 
0.915X10-3 

000 

0.33 

(  72 
J    96 
1  120 
L144 

0.457X10-3 
0.437X10-3 
0.421X10-3 
0.408X10-3 

2.12X10-3 
2.12X10-3 
2.11X10-3 
2.11X10-3 

0.420X10-3 
0.403X10-3 
0.390X10-3 
0.381X10-3 

0.960X10-3 
0.953X10-3 
0.946X10-3 
0.941X10-3 

00 

0.416 

{  72 
J    96 
1  120 
U44 

0.556X10-3 
0.536X10-3 
0.517X10-3 
0.504X10-3 

2.15X10-3 
2.14X10-3 
2.13X10-3 
2.13X10-3 

0.497X10-3 
0.482X10-3 
0.468X10-3 
0.456X10-3 

0.995X10-3 
0.989X10-3 
0.980X10-3 
0.972X10-3 

0 

0.525 

f  72 
j    96 
1  120 
1144 

0.678X10-3 
0.653X10-3 
0.633X10-3 
0.615X10-3 

2.18X10-3 
2.17X10-3 
2.16X10-3 
2.15X10-3 

0.591X10-3 
0.569X10-3 
0.554X10-3 
0.541X10-3 

1.05X10-3 
1.03X10-3 
1.02X10-3 

1.01x10-3 

1 

0.665 

(  72 
J    96 
1  120 
tl44 

0.825X10-3 
0.791X10-3 
0.766X10-3 
0.749X10-3 

2.23X10-3 
2.21X10-3 
2.20X10-3 
2.19X10-8 

0.691X10-3 
0.670X10-3 
0.655X10-3 
0.640X10-3 

1.  10  X  lO-3 
1.09X10-3 
1.08X10-3 
1.07X10-3 

2 

0.835 

(  72 
J    96 
1  120 
1  144 

0.989X10-3 
0.945X10-3 
0.920X10-3 
0.905X10-3 

2.29X10-3 
2.27X10-3 
2.26X10-3 
2.25X10-3 

0.800X10-3 
0.779X10-3 
0.756X10-3 
0.744X10-3 

1.17X10-3 
1.16X10-3 
1.14X10-3 
1.14X10-3 

*  This  table  can  also  be  used  for  single-phase  lines. 


208 


FORMULAE  AND  TABLES  FOR  THE 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    209 


210 


FORMULAE  AND  TABLES  FOR  THE 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    211 


_  CO  *O 

CO*          CO*  CO*         CO*          03 


CO*         3         M*          0*         N*          N*          S         S          3         3          3         S 


OT    X 


212  FORMULAE  AND  TABLES  FOR  THE 

The  general  solution  of  equations  (2)  giving  the  current  and 
voltage  at  any  point  on  the  line  is  as  follows: 


i--, 

or 

I  =  A^+A2e^  ) 

E=(a,-  joj  (-  Altrv*  +  A#v*)t  J 
where 


co  +  g      Cja  +  g' 

(14) 


02  = 

The  quantity  01  —  jaz  is  frequently  expressed  in  the  following 
form: 

fli      yaa_     ^       __V(Ljco  +  ^)(Q-co  +  6r)  =  VL^  +  ^    (15) 


and  is  sometimes  denoted  by  ZQ.  The  quantity  Z0  is  called  the 
initial  sending-end  impedance. 

Equations  (12)  are  to  be  used  when  the  distance  is  considered 
from  generator  end,  in  the  direction  of  decreasing  power,  and 
equations  (13)  are  to  be  used  when  the  distance  is  considered 
from  the  receiving  end,  in  the  direction  of  increasing  power. 

The  solution  of  equations  (2)  may  also  be  expressed  in  terms 
of  hyperbolic  functions. 

7  =  AI  cosh  Vs  +  A2  sinh  Fs,  ) 

E=-(<L!  -jo*)  (A,  sinh  Vs  +  Az  cosh  7s), ) 

or  7  =  ^4.1  cosh  Vs  +  A2  sinh  Vs 

E  =  (ai  —  ,702)  (Ai  sinh  Vs  +  A2  cosh  Vs) 


j  (17) 


Equations  (16)  are  to  be  used  when  the  distance  is  considered 
from  the  generator  end  and  equations  (17)  when  the  distance  is 
considered  from  the  receiving  end.  The  two  sets  of  equations 
(12),  (13)  and  (16),  (17)  are  equivalent. 

The  development  of  the  application  of  hyperbolic  functions 
to  the  study  of  electrical  problems  is  mainly  due  to  Prof.  A.  E. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    213 


Kennelly.     His  researches  on  this  subject  were  published  in  a 
number  of  very  interesting  papers.* 

The  values  of  the  constants  Ai}  A2  in  equations  (12),  (13),  (16) 
and  (17)  are  to  be  determined  from  the  boundary  conditions. 
The  evaluation  of  the  constants  in  most  cases  and  the  rearrange- 
ment of  the  final  solution  so  as  to  put  it  in  the  simplest  form  is 
a  somewhat  laborious  algebraic  process.  We  shall,  therefore,  in 
accordance  with  the  general  plan  of  this  work,  eliminate  all  the 
intermediate  steps  in  the  work  here,  and  just  give  the  final  solu- 
tions for  some  of  the  important  cases  which  occur  in  practice. 

INFINITE  LINE 

The  simplest  case  is  that  of  an  infinite  line  because  there  is  no 
reflected  wave  from  the  receiving  end.  Let  E0  be  the  generator 
voltage  assumed  to  be  of  sine  wave  form.  Then  the  voltage  and 
current  at  any  point  distance  s  from  generator  end  are 

E  =  Ege-aa  cos  0s, 


/  =  -7=^=  <r«s  cos  03s  -  0) 


(18) 


Tjl 

The  ratio  •=• 


+  «22  and  is  the  same  for  every  point  on  the 
line,  that  is,  the  line  acts  simply  as  an  impedance  whose  value  is 


a2 


The  current  leads  the  voltage  by  an  angle  0  whose  value  is 


(19) 


and  which  is  constant  for  all  points  on  the  line. 

If  g  =  0  and  Lo>  =  0,  which  is  approximately  the  case  in  cables 
in  which  the  dielectric  losses  and  the  inductance  are  small,  we 
have 


(20) 


0  =  tan"1  -  =  tan"1 1  =  45°. 
*  References  are  given  at  the  end  of  this  chapter. 


214 


FORMULAE  AND  TABLES  FOR  THE 


Expressed  in  hyperbolic  functions  the  formulae  for  the  current 
and  potential  in  an  infinitely  long  line  are 


-p 

I  = 2-r—  (sinh  Vs  —  cosh  Vs), 

ai  -  j<h v 

E  =  Eg  (cosh  Vs  -  sinh  Vs). 


(21) 


Since  V  =  a  +  j(3  is  a  complex  quantity,  cosh  Vs  and  sinh  Vs  are 
complex  quantities.  If  we  should  separate  them  into  their  real 
and  imaginary  components,  equations  (21)  would  reduce  to  the 
same  form  as  equations  (18). 

LINE  OF  FINITE  LENGTH  AND  OPEN  AT  RECEIVING  END 

Denoting  by  Eg  the  impressed  voltage  and  by  I  the  length  of 
the  line,  the  voltage  and  current  at  any  point  on  the  line  distant 
s  from  the  end  of  the  line  are 


Eg  j  cos  g 


cos 


-  «•)  +  j  sin  jte  (eas  -  <r  as]  \ 
-«0  +  j  sin  #  (e^  -  e~al) 


Eg     (  cos  ffs  -(e**  -  €-"*)  +  j  sin  ffs  (ea 


*  +  e~as)  ) 
-  e~«0  -J  ' 


(22) 


cos  # 


The  absolute  values  of  E  and  /  at  any  point  on  the  line  are 


2  COS  2  # 


(23) 


At  the  end  of  the  line,  that  is  s  =  0, 

Er  =  -7=== 


At  the  generator  end  of  the  line,  s  =  I, 
Ia  = 


E0 


VV  +  a22  v  € 


\h— 

V62«Z 


+  €-2«i  -  2  cos  2  (31 


(24) 


(25) 


The  solution  for  the  distribution  of  the  current  and  potential 
along    a  finite  line  open  at  the  receiving  end,  in  terms  of  hyper- 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    215 


bolic  functions,  and  considering  the  distance  from  the  receiving 
end  of  the  line,  is  as  follows: 

j          Eg       sinh  Vs 


—  jaz  cosh  VI 
cosh  Vs 
'coshFZ* 


(26) 


We  have 


cosh  VI 
^tanhF/. 


(27) 


Formulae  (22)  and  (25)  are  identical,  and  either  can  be  converted 
into  the  other.     This  can  be  best  shown  by  a  numerical  example. 

Let  us  take  the  same  data  as  given  in  example  worked  out  on 
page  202. 
L  =  2.21mh.,    C  =  0.0135mf.,   72  =  13.3  ohms,  0  =  0,   Z=150miles. 

For  the  frequency  of  1000 

a  =  8.72  X  10~3,    ft  =  10.3  X  10~3,     Ceo  =  84.8  X 


e2al  =  13.69,     €~2al  =  0.073,     cos  2  ftl  =  -  0. 
By  formulae  (24)  and  (25)  we  get 

2Ea 


V  13.69 +  0.07-  1.996 

En 


V13.69  +  0.07  +  1.996 


V (121)2  +  (103)2  V13.69  +  0.07  -  1.996 


=  0.007 


To  use  the  hyperbolic  function  formulae  we  must  obtain  the  values 
of  sinh  VI  and  cosh  VI. 
We  have 

sinh  VI  =  sinh  (a  +  j$)l  =  sinh  oil  cos  ftl  +  j  cosh  al  sin  ftl, 
cosh  VI  =  cosh  («  +  jft)l  =  cosh  al  cosh  ftl  +  j  sinh  al  sin  ftl, 
smhal  =  sinh  1.308  =  1.714, 
cosh  a/  =  cosh  1.308  =  1.985; 
hence, 

sinh  VI  =  1.714  cos  ftl  +j  1.985  sin  0Z, 
cosh  VI  =  1.985cos#+.7*1.714sin#Z. 


216  FORMULAE  AND  TABLES  FOR  THE 

By  (27)  we  get 


1.985  C06#  +j  1.714  sin  01 
The  absolute  value  is 


Er 


Ef. 


0.58  E0 


V  (1.985)2  cos2#  +  (1.714)2  sin2/?/ 
Eg  (1.714  cos  #  +  j  1.985  sin  0Z) 
~  (01  ~  ja*)  (1-985  cos  0Z  +  j  1.714  sin  #) 

V(1.714)2cos2#  +  (1.985)2sin2/3Z 


-JR 


V  J  (121)2  +  (103)2i  I  (1.985)2  cos2/3Z  +  (1.714)2  si 


=  0.007  Eg. 

The  agreement  in  the  results  obtained  by  the  two  methods  is 
exact.  For  this  particular  case  formulae  (24)  and  (25)  are  more 
suitable  for  numerical  calculations. 

SHORT  LINES 

For  short  lines,  where  al  is  small,  we  may  put  eal  =  e~  <*  =  1  and 
equations  (23)  reduce  to 

COS0S 

(28) 


(29) 


/.  = 


By  equation  (14)  we  have 

2,       ,  _ 


Introducing  the  values  of  a  and  0,  equation  (29)  reduces  to 


If  we  neglect  resistance  and  leakage  we  have 

a  2  .  aj=L. 
C 

Hence  equations  (28)  become 


g-fi 


cos  0s 


cos 


sin 


L  cos  j8Z 


(30) 


(31) 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    217 
Hence  *  =  y/| 


The  quantity  V  ^  is  called  the  natural  impedance  of  the  line. 


LINE  GROUNDED  AT  RECEIVING  END 

Assuming  here  also  that  we  have  a  sine  e.m.f.  of  amplitude  Eg 
impressed  on  the  line  at  the  generator  end,  we  have  the  follow- 
ing expression  for  the  current  and  voltage  on  the  line,  consider- 
ing distance  from  receiving  end. 


E  = 


Eg  cos  ffs  (ea<  +  €"«')  +  j  sin  0s  (e?8  - 
ai  -  jdz  cos  /3Z  (tal  -  e-al)  +  j  sin  fil  (eal  -f 
Eg  I  cos  /3s  (ea«  —  €~a8)  +  j  sin  0s  (eas  +  €-<") 


The  absolute  values  of  I  and  E  are 
j  _ 

E  = 


a2 


J*E. 

\    €2al 


2  COS  2  j8 


-2fl(i  -  2  cos  2  |8Z 


-2«8  -  2  cos  2  0s 


The  currents  at  the  generator  and  receiving  ends  are 


-\-e~2al  -  2  cos- 2$ 


(32) 


(33) 


(34) 


(35) 


Developing  the  same  problem  by  the  use  of  hyperbolic  functions 
we  obtain  the  following  formulae: 


Eg  cosh  7s 
(0,1  —  jdz)  sinh 
Eg  sinh  7s 
sinh7Z 


(36) 


218  FORMULAE  AND  TABLES  FOR  THE 

The  currents  at  the  generator  and  receiving  ends  are 

E  cosh  VI 


7     _     ^O 

J-r    /  \        •       i       TT7 

\#i  —  jdv)  sinh  V I 

/,      i 


sinhFr 


- 

cosh 


sech  VI  (38) 


As  an  illustration  we  may  use  the  same  data  as  given  in  the 
problem  on  page  215. 

c2«<  =  13.7,    e-2«<  =  0.07,    cos  2  01  =  -  0.998,    <n  =  121,    a2  =  103. 
sinh  VI  =  1.714  cos  01  +  j  1.985  sin  01. 
cosh  7Z  =  1.985  cos/3/  +  j  1.714  sin0Z. 

By  (34)  we  get 


/.  -    7         *'  V13.7  + 0.07 -1.996  = 

V(121)2  +  (103)2  V13.7  +  0.07  +  1.996 

Ir  =    ,  2fg  =  0.0037^,. 

V(121)2  +  (103)2  V13.7  +  0.07  +  1.996 

Working  out  this  example  by  the  use  of  formulae  (37)  gives  iden- 
tical results  which  can  be  seen  by  inspection. 


LINE   SHORT-CIRCUITED  AT  RECEIVING  END 

A  sine  e.m.f.  of  maximum  amplitude  Eg  is  impressed  on  one  end 
of  the  line,  and  the  other  end  is  short  circuited.  Considering 
distance  from  receiving  end,  that  is,  short-circuited  end,  we  have 
the  following  expressions  for  the  current  and  voltage  distribu- 
tion on  the  line: 

,=        Eg  \  cos  0s  (€?s  +  e-a*)  +  j  sin  0s  (eas  -  €~as)  j 

~~  (ai  —  ja2) \ cos 01  (eal  —  €~aO  +  j  sin 01  (eal  +  €~al)  \ 

(39) 
-,      Eg  \  cos  0s  (eas  —  e~as)  +  j  sm  0s  (eas  +  €~as)  j 


cos  01  (f?1  -  €~al)  +  j  sin  01  (ea<  +  €~aO 
The  absolute  values  of  /  and  E  are 


7  =         Eg  Ve*a*  +  €~2as  +  2  cos  2  0s 

Vai2  +  «22  Ve2"'  +  €~2al  —  2  cos  2  01 
_  Eg  V~^~+  e-*a8  -  2  cos  2  0s 
Ve2al  +  €-2aZ  -  2  cos 2/S 


(40) 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    219 
and  for  s  =  I, 


j.        _                        EgVf?«l-t 

-  €~2al  +  2  cos  2  #                       ai  ^ 

VaS  +  «22  Ve2 

at  _f_  €-2ai   _  2  COS  2  j8Z 

For, 

'  =  0, 

Tr  - 

2^ 

VaJ  +  a22  Ve2 

a/  +  c-2aZ   _  2  COS  2  01 

(42) 

Er=0. 

Ir                                         2 

•                               (43) 

+  €-2aZ+2cos2^ 

Developing  the  same  problem  by  the  use  of  hyperbolic  functions 
we  obtain  the  following  formulae: 


E  = 


For  s  =  I, 

For  s  =  0, 


/„  = 


Eg  cosh  Vs 
(ai  —  joz)  sinh  VI 
Eg  sinh  7s 
sinh  VI 

Ea  coth  VI 


cosh  FZ 


=  sech  VI. 


(44) 


(45) 


(46) 


(47) 


Formulae  (44)  to  (47)  are  the  exact  equivalents  of  formulae  (39) 
to  (43) 

POWER  TRANSMISSION 

In  the  case  of  power  transmission  the  load  conditions  at  the 
receiving  end  must  be  taken  into  consideration.  The  problem  as 
it  frequently  presents  itself  is  as  follows:  Given  the  voltage  and 
current  either  at  generator  or  receiving  end,  .to  determine  the 
values  of  voltage  and  current  at  the  other  end  of  the  line. 


220  FORMULAE  AND  TABLES  FOR  THE 

We  shall  first  consider  the  case  when  the  voltage  and  current 
at  the  receiving  end  are  known. 

Let  Er  =  voltage  at  receiving  end, 

Ir  =  current  at  receiving  end. 

Ir  =  Ir  ±  jlr"  (a  complex  quantity),  giving  the  amplitude  of  the 
current  as  well  as  its  phase  relation  with  respect  to  Er.  The 
plus  sign  is  to  be  used  when  the  current  is  leading  and  the  minus 
sign  when  the  current  is  lagging. 

Considering  the  distance  s  from  receiving  end,  we  have  the  fol- 
lowing expressions  for  the  voltage  and  current  at  any  point  on 
the  line: 

I  =  i  IKJS  =F  KJr"  +  K2Erp  - 


(48) 


where 

K2  =  (««  -  c-«8)  cos /3s,  (49) 

K3  =  (eaa  +  e-as)sin/3s, 

II  «2 


=  \  [KiEr+  K2  (ajrf  d=  ajr")  +  Kz  (aJS  T  ajr")  } 

K,  (±  ajr"  -  02//)  +  K3  (ajrf  ±  02//0  i  , 


P  =  — 2 o,        g  = 

The  upper  sign  before  //'  in  equations  (48)  is  to  be  used  when 
the  receiving  current  is  leading  the  voltage,  and  the  lower  sign 
is  to  be  used  when  the  receiving  current  is  lagging. 

To  get  Eg  and  Ig,  the  voltage  and  current  at  generator,  it  is 
necessary  only  to  put  s  =  I'm  equations  (49). 

If  we  put  equations  (48)  in  the  form 


the  absolute  values  of  7  and  E  are  given  by 

(50) 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS  221 

The  phase  relations  between  currents  and  voltages  at  receiving 
and  generator  ends  are  as  follows: 

Let    0i  =  phase  angle  between  current  and  voltage  at  receiving 
end. 

02  =  phase  angle  between  current  at  generator  end  and 

voltage  at  receiving  end. 

03  =  phase  angle  between  voltages  at  generator  and  re- 

ceiving ends. 


I 

0!  =        -1- 


I  " 

02  =  tan-1-^-, 

LQ 
Tji   It 

A  =  tan-'b 


(51) 


If  the  values  of  the  tangents  of  0i,  02  and  03  are  positive  we  have 
leading  phase  angles,  and  if  the  values  are  negative  we  have 
lagging  phase  angles. 

We  have  expressions  similar  to  those  of  (48)  to  (50)  when  the 
current  and  voltage  are  given  at  the  generator  end.  Let  Eg 
denote  the  voltage  at  generator  end,  Ig  the  current  (in  intensity 
and  in  phase  with  respect  to  Eg)  at  the  generator  end,  that  is, 
Ig  =  Ig'  db  jig" -  Considering  the  distance  s  in  this  case  from 
the  generator  end  we  have 


(52) 


The  values  of  KI,  K2,  K-6)  Kt  are  given  by  equation  (49).  The 
double  signs  before  Ig"  have  the  same  significance  as  in  equation 
(48).  The  upper  sign  is  to  be  used  when  the  generator  current  is 

I  " 

leading  the  generator  voltage  by  the  angle  whose  tangent  is  -f-r 

*g 

and  the  lower  sign  is  to  be  used  when  the  current  is  lagging. 

Equations  (48)  to  (52)  give  all  the  formulae  needed  in  work- 
ing out  most  problems  in  long-distance  power  transmission. 


I  -  \  {KJgf  =F  KJg"  -  K2Egp 

+ 1  { ±  KJg"  +  KJg'-  K2Egq  -  K>Egp\, 
E  =  }  {KJEg  -  K2  (ajaf  ±  aJ0")  -  K3  (a2//  T  ajg 
K2  (02//  =F  ai//0  -  K3  (a,Igf  db 


222  FORMULAE  AND  TABLES  FOR  THE 

HYPERBOLIC  FORMULA 

Another  set  of  formulae  corresponding  to  those  of  (48)  to  (52) 
has  been  worked  out  in  terms  of  the  hyperbolic  functions. 

If  the  electrical  conditions  are  given  at  the  receiving  end, 
considering  the  distance  s  from  receiving  end  we  have 

7  =  Ir  cosh  Vs  +  Er(p  +  jq)  sinh  Vs,    )  /^x 

E  =  Er  cosh  Vs  +  Ir  (dt  -  ja»)  sinh  7s. ) 

If  Eg  and  Ig  are  given,  that  is,  the  current  and  voltage  at  generator 
end,  considering  the  distance  s  from  generator  end  we  have 

I  =  Ig  cosh  Vs  —  Eg  (p  +  jq)  sinh  Vs,     )  ^A\ 

E  =  E0  cosh  7s  -  Ig  (ai  -  joj)  sinh  7s.  ) 

Formula  (53)  corresponds  to  formula  (48)  and  formula  (51)  corre- 
sponds to  formula  (52),  the  various  terms  having  the  same  signifi- 
cance as  in  the  above  equations.  The  terms  sinh  7s  and  cosh  7s 
are  complex  quantities.  They  can  be  put  in  the  form  of  two 
terms,  a  real  and  an  imaginary,  thus, 

sinh  7s  =  sinh  (a  +  ,//3)  s  =  sinh  as  cos  /3s  +  j  cosh  as  sin  /3s,  1 
cosh  7s  =  cosh  (a  +  j]8)  s  =  cosh  as  cos  0s  -j-  j  sinh  as  sin  /3s.  ) 
The  values  of  sinh  as  and  cosh  as  can  be  obtained  from  a  table 
of  exponential  functions. 

A  short  table,  Table  XX,  is  also  appended  at  the  end  of  this 
chapter  giving  directly  the  values  of  cosh  7s  and  sinh  7s.  The 
range  of  values  of  7s  covered  by  this  table  is  sufficient  for  all 
practical  problems  in  power  transmission. 

Formulae  (53)  and  (54)  appear  simpler  and  more  compact  than 
the  corresponding  formulae  (48)  and  (52),  but  the  labor  involved 
in  numerical  computation  by  the  hyperbolic  formulae  is  not  any 
less  than  by  the  other  formulae,  and  there  is,  therefore,  little  choice 
in  regard  to  the  use  of  the  two  different  sets  of  formulae  for  numeri- 
cal problems.  It  is  desirable,  however,  to  be  able  to  check  the 
results  by  two  different  methods,  so  that  any  possibility  of  numeri- 
cal errors  may  be  eliminated. 

APPROXIMATE  FORMULAE  FOR  SHORT  LINES 

If  the  electrical  conditions  are  given  at  the  receiving  end, 


=  Ir  \  1  +*>(*'2    y)+jtt/fe*|  +  Er  \  (p  +  jq)  (a 


(56) 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    223 
If  the  electrical  conditions  are  given  at  the  generator  end, 


o2  (^2  _   OZ\  )  ^       ' 

-JVf  1 1  +      ^ %    ^  ^  +  j«/3s2 (  -/, { (ai  - jaO  (a  +#)«{. 
(  ^  )  I 

The  above  two  formulae  were  given  by  W.  E.  Miller.*  He  states 
that  they  can  be  used  with  an  accuracy  of  1  per  cent  for  lines 
using  No.  2  wire  up  to  120  miles  at  60  cycles  and  150  miles  at 
25  cycles.  Greater  accuracy  is  obtained  if  larger  wires  than 
No.  2  are  used,  though  the  difference  is  immaterial. 

•\ 
ILLUSTRATIONS 

Example.  —  The  following  example  is  taken  from  Mr.  Miller's 
paper.  We  shall  work  it  out  by  the  formula  (48)  and  compare 
the  results  with  those  obtained  by  Mr.  Miller  by  the  use  of  the 
hyperbolic  method. 

Three-phase  line,  300  miles  long  using  hard-drawn  stranded 
copper  wire,  No.  000  B.  &  S.  triangularly  spaced,  with  wires  10  ft. 
apart.  Frequency,  60  cycles.  At  the  receiving  end  we  have  the 
following  conditions.  Line  voltage  104  kv.  or  60  kv.  between  wire 
and  neutral;  load  current  100  amperes  at  receiving  end  at  90 
per  cent  power  factor  lagging.  Then 
Er  =  60  kv., 

IT  =  IT  -  jlr"  =  90  -  j  43.5  amperes, 
R  =  0.33  ohm  per  mile,  L  =  2.13  mh.  per  mile, 

C  =  0.014  mf.  per  mile,  g  =  0  (neglected), 

a  =  0.421  X  10~3,  0  =  2.11  X  10~3, 

al  =  0.126,  01  =  0.633  =  36°  16'  5", 

e«<  +  e-ai  =  2.016,  cos  0Z  =  0.8062, 

2*1  -  e-«i  =  0.253,  sin  01  =  0.5915, 

Ki  =  1.625,      K2  =  0.204,      K3=  1.193,       KI  =  0.149, 
ai  =  392,       a2  =  78,        p  =  2.44  X  10~3,     q  =  0.485  X  10~3. 

Introducing  these  values  in  equation  (48)  we  get  for  current  and 
voltage  at  generator  end 

Ig  =  73.94  +j  61.65  amp., 

Eg  =  66.4 +.7  21. 04  kv. 

*  W.  E.  Miller,  Formulae,  Constants  and  Hyperbolic  Functions  for  Trans- 
mission Line  Problems,  General  Electric  Review  Supplement,  May,  1910. 


224  FORMULAE  AND  TABLES  FOR  THE 

The  absolute  values  of  current  and  voltage  are 

/„  =  V  (73.94)2  +  (61.65)2  =  96.3  amp. 
Ea  =  V(66.4)2  +  (21.04)2  =  69.6  kv. 

The  generator  current  leads  the  voltage  at  the  receiving  end  by 
an  angle 

=  tan-1  0.834  =  39°  48'. 


The  generator  voltage  leads  the  voltage  at  the  receiving  end  by 
an  angle 


tan-1  =  tan-1  0.317  =  17°  36'. 

DO.OO 

The  current  at  the  generator  end  leads  the  generator  voltage  by 
the  angle 

(39°  48'  -17°  36')  =  22°  12'. 

The  power  factor  at  the  generator  end  is 

cos  (22°  120  =  0.926. 
Transmission  efficiency, 

60  X  100  X  0.9 
13      69.6  X  96.3  X  0.926 

To  get  the  value  of  the  charging  current  put  Ir  =  0,  no  load  current. 
Then          I0  =  -  2  A3  +  j  90.3  =  90.3  amperes  per  wire. 

The  values  obtained  by  Mr.  Miller  for  this  example  using  the 
hyperbolic  functions  are 

Ig  =  74.6  +J61.9  amp.,         Eg  =  66.2  +  j21.0kv. 

The  values  differ  very  slightly  from  the  values  obtained  by 
formula  (48)  and  even  this  slight  difference  may  be  accounted  for 
by  the  fact  that  in  using  the  hyperbolic  tables,  interpolations  are 
required,  and  apparently  Mr.  Miller  only  used  first  differences. 

It  is  interesting  to  obtain  the  generator  current  and  voltage  and 
the  .different  phase  angles  for  different  load  currents.  Since  we 
have  given  in  detail  the  numerical  work  for  the  case  of  100  am- 
peres load  current,  it  will  be  sufficient  to  simply  tabulate  the 
results  for  other  load  currents. 


CALC  ULA  TION  OF  ALTERNA  TING  C  URRENT  PROBLEMS    225 


/,. 

25 

50 

100 

150 

Ig'+jlg" 

16  7+7  83  1 

35  .9+j  74  .9 

73.  9+/  61.  65 

112.  l+j  47.  3 

Eg'+jEg"  

53.2+.;  8.  6 

57.9+j  12.6 

66.  4+;  21.  04 

75.  2+;  29.  3 

V7,'2  +  7/"  

84.8  amp. 

83  amp. 

96.3  amp. 

121  .  7  amp. 

VEa'2  +  Ea"2 

53  8kv 

59  2  kv 

69  6kv 

80  7kv 

Phase  angle,  IgEr    j 
Phase  angle,  EgEr   j 

Phase  angle,  I0Eg    < 

rj  =  efficiency    of     ( 
transmission  .  .  .  .  ( 

78°  40' 
leading 
9°  12' 
leading 
69°  28' 
leading 

84% 

64°  18' 
leading 
12°  18' 
leading 
52°  leading 

89.2% 

39°  48' 
leading 
17°  36' 
leading 
22°  12' 
leading 

88% 

22°  54 
leading 
21°  16' 
leading 
1°38' 
leading 

82.5% 

As  an  illustration  of  the  application  of  formulae  (52)  and  (54) 
we  shall  take  the  following  example: 

A  three-phase  line  made  up  of  No.  0000  B.&  S.  copper-strand 
wire  triangularly  spaced  with  10  feet  between  wires;  frequency, 
60  cycles;  length  of  line  200  miles. 

Voltage  at  generator  between  wires,  104  kv.,  between  wire  and 
neutral,  60  kv. 

Ig  =  120  amperes  at  .90  per  cent  power  factor  leading. 
I0  =  108  +j  52.3, 

Eg    =    60. 

To  determine  the  voltage  and  current  at  receiving  end  and  the 
various  phase  angles. 

R  =  0.26  ohm  per  mile. 
L  =  2.09  mh.  per  mile. 
C  =  0.0143  mf.  per  mile. 
a  =  0.344,     ft  =  2.10  X  10~3. 
01  =  0.420  =  24°  3'  50", 
cosjSZ  =  0.913,     sin/3Z  =  0.408, 
(?i  +  €-«i  =  2.005,     €?1  -  €-<*'  =  0.1376, 
K!  =  1.831,    K2  =  0.126,     K3  =  0.818,    K^  =  0.0561, 
ai  =  396,     02  =  62.5, 
p  =  2.47  X  10~3,     q  =  0.39  X  1Q-3. 

Introducing  the  above  values  in  equations  (52)  we  get 

Ir  =  97.64- .7 11. 18  amp., 
Er  =  57.74  -j  18.02  kv. 


226  FORMULAE  AND  TABLES  FOR  THE 

Using  formulae  (54)  for  this  example,  we  find  from  Table  XX  by 
interpolation 

cosh  VI  =  cosh  (0.0688  +  ,7  0.42)  =  0.915  +j  0.028, 
sihh  VI  =  sinh  (0.0688  +j  0.42)  =  0.063  +j  0.409, 


and  using  the  values  of  p,  q,  ax  and  a2  as  obtained  above  we  get 

Ir  =  97.6-  j  11.19. 
Er  =  57.7  -  j  17.98. 

The  agreement  in  the  values  of  Ir,  Er  as  calculated  by  the  two 
different  formulae  is  very  close  indeed. 

The  absolute  values  of  the  current  and  voltage  are 

Ir  =  V(97.6)2  +  (11.2)2  =  98.2  amp., 
Er  =  V  (57.7)2  +  (18)2  =  60.4  kv. 

The  current  at  receiving  end  lags  behind  voltage  at  generator 

end  by  angle  tan-1  ^|  =  tan-1  0.101  =  5°  46'. 
y  i  .o 

The  voltage  at  receiving  end  lags  behind  voltage  at  generator 

i  o  r»o 

end  by  angle  tan'1  ^^  =  tan"1  0.312  =  17°  20'. 
o/  .  / 

Therefore,  the  current  at  receiving  end  leads  the  voltage  at 
receiving  end  by  the  angle 

17°  20'  -  5°  46'  =  11°  34'. 

,_  ffi  .  98.2  X  60.4  X  0.98      n  on 

Transmission  efficiency  =    12Q  x  60  X  0  90     =          per         ' 

GENERATOR    VOLTAGE    AND    IMPEDANCE    AT    RECEIVING    END 

KNOWN 

The  problem  in  long  distance  transmission  may  also  present 
itself  in  the  following  form:  Given  the  potential  at  the  generator 
end  and  the  impedance  at  the  receiving  end,  to  determine  the 
current  at  the  generator  end  and  the  potential  and  current  at  the 
receiving  end. 

Let  Eg  denote  the  voltage  at  the  generator  end, 

Z  denote  impedance  at  the  receiving  end. 

Distance  is  considered  from  receiving  end. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    227 
The  values  of  Ig,  Er,  and  7r  are  given  by  the  following  formulae: 


Ir  = 


(q-jh)(c  +  l+jd) 


/„  = 


E,(K1+jKt)(a1+jaa)  (q  -  jh) 


(58) 


where 


"'-oT  a2  =  c^' 

e+*-J^S±£-  <»> 

q  =  ctal  cos  01  —  d(?1  sin  01  —  €~al  cos  01, 

h  =  Cf?1  sin  01  +  rfeaZ  cos  01  +  €~ai  sin  /?/, 

KI  =  ceal  cos  j8/  —  deal  sin  j8Z  +  €~aZ  cos  /3Z, 

-^2  =  ceal  sin  /3/  +  c?€aZ  cos  01  —  e~al  sin  01. 

In  terms  of  hyperbolic  functions  we  obtain  the  following  set  of 
formulae  for  this  problem: 


Eg  jcoshFs  +      si 
/  Z° 


E  _ 


Z0  sinh  VI  +  Z  cosh  VI 
_Eg\ZQ  sinh  Vs  +  Z  cosh 


Z0  sinh  VI  +  Z  cosh 


(60) 


where  Z0  has  been  put  for  brevity  in  place  of  ai  —  ^2. 

For  /  and  E  at  the  receiving  end,  we  put  s  =  0  in  formulae  (60) 
and  we  get 


lr 


Z0  sinh  VI  +  Z  cosh  VI 

E  =  E°Z 

ZosmhVl  +  ZcoshVl' 

At  the  generator  end,  s  =  I,  we  have 

>i+z 

+  Z~o 


(61) 


Z0  sinh  VI +  Z  cosh  FZ 


(62) 


228  FORMULAE  AND  TABLES  FOR  THE 

Example.  — 

Let        Z  =  540  +  j  261,        V(540)2  +  (261)2  =  600, 
and  let  us  take  for  the  line  constants  the  same  as  those  given  in 
the  example  on  page  223. 

R  =  0.33  ohm  per  mile, 
L  =  2.13  mh.  per  mile, 
C  =  0.014  mf.  per  mile, 

0  =  0,     1  =  300  miles,     co  =  2;rX60,     Eg  =  60  kv., 
al  =  0.126,     #  =  0.633, 

&l  =  1.134,  €~al  =  0.882,  cos/3Z  =  0.806,  sin#  =  0.5915, 
01  =  392,        02  =  78, 

392-J78  +  540+J261          1;i 
=  =-1.4 


$  =  -3.46,     h  =  1.47,    #1  =  -2.04,    K2  =  0.43, 
52  +  h2  =  14.13,     (ai2  +  022)  (S2  +  /i2)  =  2,257,240. 

Introducing  these  values  in  formulae  (58)  we  get 

Eg  (-3.46  -j  1.47)  (-2.46  +  J2.ll) 
^r  =  -  14  13  ~  =  ug  (0.82  -  j  0.26) 

=  49.2  -  j  15.6  kv. 

The  absolute  value  of  Er  is  V(49.2)2  +  (15.6)2  =  51.6  kv. 

The  voltage  at  receiving  end  lags  behind  the  generator  voltage  by 

the  angle 

tan-1^  =  tan-1  0.317  =  17°  36'. 

,  =  E0  (-3.46  -j  1.47)  (-0.46  +  J2.ll)  (392  +  j  78) 

2,257,240 
=  gg  (1.043^0.987)  =  626 


absolute  value  =  V  (62.6)2  -f  (59.5)2  =  86.4  amp. 

The  current  at  receiving  end  lags  behind  the  generator  voltage 
by  the  angle 

tan-1  ^|  =  tan-1 0.934  =  43°  3'. 

=  Eg  (-2.04  +  .7  0.43)  (392  +  j  78)  (-3.46  -j  1.47) 
0  2,257,240 


absolute  value  =  V(76.8)2  +  (31.7)2  =  83.1  amp. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    229 

The  current  at  the  generator  end  leads  the  generator  voltage  by 
the  angle 

tan-1  ?^  =  tan-1 0.413  =  22°  27'. 

7o.o 

The  current  at  the  receiving  end  lags  behind  the  voltage  at  the 
receiving  end  by  the  angle 

43°  3'  -  17°  36'  =  24°  27'. 

At  the  generator  end  the  power  factor  is  cos  (22°  27')  =  0.924  and 
at  the  receiving  end  the  power  factor  is  cos  (24°  27')  =  0.910. 

Efficiency  of  transmission, 

51.6  X  86.4  X  0.910 
77  =    60X83.1X0.924    ^percent. 

Working  out  the  same  problem  by  the  hyperbolic  formulae  we  get 
cosh  VI  =  cosh  (al  +jpl)  =  cosh  (0.126  +j  0.633) 


sinh  VI  =  sinh  (al  +J0J)  =  sinh  (0.126  +j 0.633). 
From  Table  XX  we  find 

cosh  (0. 12 +j  0.64)  =  0.808 +j  0.072, 
cosh  (0.12 +j  0.62)  =  0.820  +j  0.070, 
cosh  (0.14  +  .7  0.64)  =  0.810  +  j  0.083, 
cosh  (0.14+  j  0.62)  =  0.822  +  j  0.081 . 

Therefore, 

cosh  (0.12  +  j  0.633)  =  0.812  +j  0.072, 
cosh  (0.14  +j  0.633)  =  0.814  +  j  0.082, 

and  cosh  (0.126+ j  0.633)  =  0.812  +  j  0.075. 

In  a  similar  way  we  obtain  from  the  same  table  by  interpola- 
tion 

sinh  (0.120  +j  0.633)  =  0.102  +  j  0.597, 
ZQ  =  ai  -  joj  =  392  -  j  78,    Z  =  540  +  j  261,    #,  =  60  kv. 

Introducing  the  above  values  in  equations  (61)  and  (62)  we  get 

".  60,000  

r~~  (392  -j  78)  (0.102  +  j  0.597)  +  (540  +  j  261)  (0.812 \+j  0.075) 

60,000 

505.4  +j  478.5 
_  60,000  (505.4  -  j  478.5) 

484,391  bAe      J  59.3  amp. 


230  FORMULAE  AND  TABLES  FOR  THE 

0.812  +j  0.075  +  ^"jj7'.2^1  (0.102  +j  0.597)  | 


Ia  = 


tfa{  0.812  +j  0.075  +(1.198  +  j  0.904  )  (0.1  02  +j  0.597)1 
(392  -  j  78)  (0.102  +  j  0.597)  +  (540  +  j  261)  (0.812  +  j  0.075) 
_  60,000  (0.394  +j  0.882)      7fi7 
505.4  +j  478.5      "  =  76'7 


505.4  +  7  478.5 


The  results  obtained  by  the  two  different  sets  of  formulae  are  'in 
exact  agreement. 

TRANSIENT  PHENOMENA  IN  CIRCUITS  HAVING  DISTRIBUTED 
INDUCTANCE  AND  CAPACITY 

In  Chapter  V  we  gave  a  brief  discussion  of  transient  electrical 
phenomena.  We  have  shown  that  if  the  circuit  conditions  are 
disturbed  in  any  way  the  current  and  potential  do  not  pass  over 
instantly  from  one  permanent  value  to  another;  it  always  re- 
quires a  certain  time  interval  before  the  current  and  potential 
reach  their  steady  values  again.  During  the  transition  period  the 
current  and  potential  may  be  considered  as  consisting  of  two 
components,  one  of  which  corresponds  to  the  permanent  value, 
and  the  other  the  transient  term  which  disappears  more  or  less 
rapidly. 

In  the  formulae  given  in  Chapter  V  we  have  considered  only 
circuits  of  localized  inductance  and  capacity,  so  that  the  space 
element  does  not  enter  into  the  formulae.  We  were  concerned 
only  with  the  time  element,  the  time  rate  at  which  the  transient 
term  disappears. 

In  problems,  however,  connected  with  long-distance  transmis- 
sion lines,  the  distributed  character  of  the  inductance,  capacity 
and  resistance  must  be  taken  into  consideration.  If,  for  instance, 
an  e.m.f.  is  impressed  on  a  long  line,  every  element  of  the  line 
is  not  charged  to  the  same  potential  at  the  same  instant.  It 
requires  a  certain  length  of  time  before  the  entire  line  is  com- 
pletely charged,  and  during  that  interval  the  charge  and  poten- 
tial differ  for  different  points  on  the  line.  In  the  discussion, 
therefore,  of  transient  phenomena  on  lines  we  must  deal  with  two 
variables,  distance  and  time. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    231 

It  is  to  be  regretted  that  the  formulae  for  transient  phenomena 
on  transmission  lines  cannot  be  presented  in  the  same  manner  as 
those  given  in  other  chapters.  The  formulae  are  generally 
very  long  and  involved,  and  if  the  bare  formulae  are  given  leaving 
out  the  derivation  and  the  complete  discussion  of  the  physical 
significance  of  the  various  factors,  their  usefulnesss  is  greatly 
reduced.  The  formulae  for  transient  phenomena  occurring  on  a 
transmission  line  have  their  greatest  value  in  enabling  us  to  get  a 
proper  appreciation  of  the  phenomena  occurring  on  the  line  irre- 
spective of  magnitude. 

It  was  thought  advisable,  therefore,  not  to  attempt  to  cover 
thoroughly  this  branch  of  the  subject,  but  to  give  only  a  few 
formulae  for  the  cases  of  most  common  occurence.  Those  inter- 
ested in  the  subject  can  refer  to  Professor  Steinmetz's  book 
" Transient  Electric  Phenomena  and  Oscillations"  where  the 
whole  subject  is  discussed  in  a  masterly  manner. 


TRANSIENT  PHENOMENA  IN  CIRCUITS  HAVING  DISTRIBUTED 
RESISTANCE  AND  CAPACITY 

Assume  the  inductance  of  the  line  negligible  as  in  the  case  of 
transatlantic  cables. 

Suppose  the  cable  is  open  at  receiving  end,  and  a  constant  con- 
tinuous e.m.f.  E  is  suddenly  applied  at  the  sending  end.  A 
charging  current  commences  to  flow  in  the  cable  and  propagates 
itself  from  point  to  point  on  the  cable  until  the  whole  cable  is 
charged  to  the  impressed  potential  E.  During  the  charging  period 
the  current  and  potential  vary  at  different  points  on  the  cable, 
so  that  the  current  and  potential  are  functions  of  the  time  and  the 
distance. 

Let    r  =  resistance  per  unit  length, 
C  =  capacity  per  unit  length, 
I  =  length  of  cable, 
E  =  impressed  e.m.f, 

s  =  distance  from  sending  end  to  any  point  on  the 
cable. 

The  potential  at  any  point  on  the  cable  distant  s  from  sending 
end  may  be  put  in  the  form, 

e=E  +  ev,  (63) 


232 

where 


FORMULAE  AND  TABLES  FOR  THE 


**  25 t 


(64) 


The  current 

2E( -?A 


COS    -  -7  H-  € 


25 1 
4r(7 


COS 


(5^s\ 
\2    l) 


(65) 


If  we  put 


Ir  =  TI  total  resistance, 
1C  =  Ci  total  capacity, 


then,  neglecting  upper  harmonics,  formulae  (64)  and  (65)  become 
approximately, 


(66) 


At  the  end  of  the  line  (s  =  I),  the  potential  is  x  per  cent  less  than 
the  maximum  at  the  time  Tx  given  by 


X 


100      x 


(67) 


As  an  illustration  we  may  take  the  cable  between  Ireland  and 
Newfoundland  which  is  2640  km.  long.  The  total  resistance  n  = 
6000  ohms,  Ci  =  40  mf . 

The  problem  is  to  determine  the  time  required  before  e  reaches 
the  value  0.95  E. 


We  have  ev  =  0.05  E,        x  =  5, 

4riCi,      400      4X6000X40X10-6,      80 


96  X  10 


-2 


X  3.27  =  0.31  second. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    233 


The  time  required  for  the  potential  to  attain  99  per  cent  of  its 
normal  value  is 


Tx  = 


4  X  6000  X  40  X  10~6  ,      400 


A  .ftn 
=  0.499  second. 


As  another  illustration  let  us  determine  the  value  of  the  potential 
and  current  for  points  100  km.,  1000  km.  and  2000  km.  distant 
at  the  time  t  =  0.1  sec.  Neglecting  the  upper  harmonics  and 
using  formula  (66)  we  have 

T2  9.87 


4  X  6000  X  40  X  10~6 
€-i.o28=  Q.358. 


10.28. 


Rise  in  Voltage  m  a  Transatlantic  Cable 


0.1  0.2  0.3  0.4  0.5      j   0.6  0.7  0.8 

FIG.  68.  —  Rise  in  voltage  in  a  transatlantic  cable. 


0.9         1.0 


For 

s  =    100 km.,     sin  (^ )  =  sin  (0.06)  =  0.06,     cos  (0.06)  =  0.998, 
l>j 


s  =  1000km.,     sinf  |  yl  =  sin  (0.60)  =  0.56,     cos  (0.60)  =  0.825, 

s  =  2000km.,     sin/!  |)  =  sin    (1.2)  =  0.93,     cos    (1.2)  =  0.362. 
Introducing  these  values  in  formula  (66)  we  get, 


8 

e 

j 

100km. 
1000  km. 
2000km. 

0.973# 
0.745  E 
0.544# 

0.7147 
0.5917 
0.2597 

234 


FORMULAE  AND  TABLES  FOR  THE 


where  /  =  — 

In  Fig.  68  a  set  of  curves  is  given  showing  the  rise  of  poten- 
tial over  the  entire  length  of  the  cable. 

In  Fig.  69  curves  are  given  for  the  current  as  a  function  of  the 
time  for  the  points  on  the  cable 

s  =  0,      s  =  0.25  I,      s  =  0.5  Z,      s  =  0.75  Z. 


Current  Distribution  on  a 
Cable  on  Sudden  Charging 


0      0.2  0.4   0.6    0.8   1.0    1.2    1.4    1.6    1.8    2.0    2.2    2.4    2.6   2.8    3.0    3.2    3.4    3.6   3.8  4.0 

t  in  Seconds 

FIG.  69.  —  Current  distribution  on  a  cable  on  sudden  charging. 


A  CHARGED   CABLE  HAVING  A   CONSTANT  E.M.F.   APPLIED   BE- 
TWEEN ONE  END  OF  THE  CABLE  AND  GROUND  AND  SUD- 
DENLY GROUNDED  AT  THE  OTHER  END 

The  permanent  values  of  the  potential  and  current  at  any  point 
on  the  line  distant  s  from  the  point  where  the  e.m.f.  is  applied  are 


-•-I 


(68) 


The  values  of  the  transient  terms  of  potential  and  current  at  any 
point  on  the  line  are  given  by  the  following  formulae: 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    235 


2E( 


1     ~ 

-         ' 


«"/f  r^s  v* '  w  «  ^ (69) 

*• =~  ?  J^'COS(T)-  «"^'«»fr) 

9  IT2  in  \  ) 

i       r"r>~  *         / "  "TT^M  r 

+  €riCl     COSf— y— )  —    •    •    •    >• 

For  the  current  at  the  end  of  the  cable  we  may  combine  the 
second  equations  of  (68)  and  (69)  and  put  them  in  the  following 
form : 

/  =  f  (l-2m-£  r^'cosrmr),  (70) 


V  m=l 

which  gives  the  arrival  curve  of  the  current  at  the  receiving  end. 
The  series  given  by  equation  (70)  is  very  slowly  convergent  for 
small  values  of  t,  and  makes  the  calculations  very  laborious. 
Equation  (70)  may  be  put  in  another  form  which  gives  a 
rapidly  convergent  series  for  small  values  of  t.  The  equivalent 
of  formula  (70)  is 


41 


As  an  illustration  we  take  the  same  data  as  given  in  the  exam- 
ple given  on  page  232. 

n  =  6000  ohms, 

Ci  =  40  mf.,         I  =  2640  km. 
For  t  =  0.001  second,  we  have 

~t  =  0.041. 
By  (70), 
7  =  ^  Jl  -  2e-°-041cos7r  -  2e-°-164cos27T  -  2  e-°-369  cos  3 TT  -  •  •  •{ 

=  -  }1  +  1.92  -  1.7  +  1.38  -••-•!. 


It  requires  a  very  large  number  of  terms  to  obtain  the   value 
of/. 


236  FORMULAE  AND  TABLES  FOR  THE 


By  formula  (71)  we  have 


_  40  X  IP"6  X  6000 
4  X  10-3 


^  -  *  X  2.9  X  e- 

which  shows  that  for  the  time  t  —  0.001  sec.  the  current  at  the 
receiving  end  is  negligible. 

If  the  cable  is  grounded  all  the  time  at  one  end  and  a  constant 
e.m.f.  is  applied  at  the  other  end  the  permanent  values  of  the 
potential  and  current  as  before  will  be 


(72) 

TI 

The  transient  terms  for  the  potential  and  current  are  given  by  the 
formulae 

•^-i          J         -^-t          >2*8  (73) 

i  C  COS  I  ~T~  I     1  *  €  COS  I  ~     ^      J    i"  *  i  * 

ri 


/27TS 

COS(— 


CIRCUITS  CONTAINING  DISTRIBUTED  RESISTANCE,  INDUCTANCE 
AND   CAPACITY;  LINE   OSCILLATIONS 

We  have  shown  in  Chapter  V  that  in  a  circuit  containing 
inductance  and  capacity,  the  transient  possesses  an  oscillatory 
character,  unless  the  resistance  exceeds  a  certain  critical  value 
depending  on  the  inductance  and  capacity  of  the  circuit.  In 
circuits  of  distributed  inductance  and  capacity,  the  transient  is,  in 
general,  oscillatory.  The  discharge  of  an  accumulated  charge  of 
atmospheric  electricity  on  a  line,  a  sudden  change  in  load,  an 
arcing  ground,  or  a  break  in  the  circuit  will  set  up  electric  oscil- 
lations on  the  line.  Owing  to  the  inductance  and  capacity  of  the 
system  a  flow  of  current  stores  up  magnetic  and  dielectric  energy 
in  the  system.  A  sudden  change  in  circuit  condition,  grounding 
the  line  or  a  break  in  circuit,  for  instance,  releases  the  energy 
stored  in  the  system  and  it  oscillates  back  and  forth  until  the 
entire  energy  is  dissipated  by  the  resistance  and  leakage  of  the 
system. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    237 

Since  the  maximum  magnetic  energy  must  equal  the  maximum 
dielectric  energy,  we  have 


-  Vi*. 


(74) 


The  quantity  Z0  is  called  the  natural  or  surge  impedance,  and  yo 
is  the  natural  or  surge  admittance. 

In  the  case  of  circuits  of  large  inductance  and  small  capacity 
even  moderate  currents  may  cause  a  dangerous  rise  in  voltage  by 
a  sudden  change  in  circuit  condition. 

The  general  expressions  for  the  oscillating  current  and  voltage 
are  as  follows: 


The  values  of  the  constants  ?„,  0n,  An,  Bn,  Pn  and  Qn  depend  on 
the  terminal  conditions,  and 


We  shall  now  consider  a  few  specific  cases. 


LINE  GROUNDED  AT  ONE  END  AND  OPEN  AT  THE  OTHER  END 

The  terminal  conditions  are 

E  =  0    when    s  =  0,     and    7  =  0    when    s  =  I 
I  is  the  length  of  the  line. 

To  satisfy  these  conditions  we  must  have 

^  7      (2  n  —  1)  TT 

Qn  =  0    and    ynl  =  -  -      J 

A 


238  FORMULAE  AND  TABLES  FOR  THE 


and  equations  (76)  reduce  to 

An  cos  pnt  +  Bn  sin  (3nt  \ , 

(78) 


cos (2  n          **  I  An  cos  ^  +  Bn  sin 


n  =  1 

_(2n-l) 


ri^~at  2) 


n-  l)27r2 


If  we  neglect  a2  under  the  radical,  as  it  is  generally  small  com- 

7T2 

pared  with  the  term  y^77  >  an(i  also  put  ZL  =  L0  and  ZC  =  C0,  so 

that  Lo  and  C0  represent  the  total  inductance  and  capacity  of 
the  line,  we  get 


j8»  is  the  frequency  constant  of  the  line. 

The  fundamental  frequency  of  the  oscillations  on  the  line  is 
given  by 

f  =  A  = 


_ 
4\/LoCo' 

We  may  therefore  write 

/3n=27r(2n-  !)/!.  (82) 

Equations  (78)  may  now  be  written  in  the  following  form: 


n  =  oo 

COS 


(83) 


The  current  and  potential  on  the  line  are  complex  waves  con- 
sisting of  fundamental  waves  and  odd  harmonics.  For  the  funda- 
mental wave  the  current  is  zero  at  the  open  end  of  the  line 
(s  =  I)  and  gradually  increases  to  its  maximum  value  at  the 
grounded  end  of  the  line.  The  potential  is  zero  at  the  grounded 
end  of  the  line  (s  =  0)  and  gradually  increases  to  its  maximum 
value  at  the  open  end  of  the  line  (s  =  I). 

The  wave  length  of  this,  the  fundamental  oscillation,  is  four 
times  the  length  of  the  line. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS  239 

The  values  of  the  constants  Dn  and  4>n  depend  upon  the  initial 
conditions,  and  have  to  be  evaluated  independently  for  each  par- 
ticular problem.  In  general  the  constants  are  determined  in  the 
usual  manner  of  evaluating  a  Fourier  Series.  We  may  best  indi- 
cate the  process  by  the  following  illustration:  Let  the  line  be 
.charged  to  a  uniform  potential  EQ  but  with  no  current  in  the  line 
before  the  discharge.  The  line  is  then  grounded  at  one  end  s  =  0 
while  open  at  the  other  end  s  =  1.  We  have  then  for  t  =  0, 1  =  0 
for  all  values  of  s  and  for 

t  =  0,    E  =  E0    for  all  values  of  s. 

From  equation  (78)  it  is  evident  that  to  satisfy  the  first  condition 
we  must  have  An  =  0,  and  to  satisfy  the  second  condition  we 
must  have 

>     •     (2n  —  I)TTS 


.., 

The  values  of  Bn  can  now  be  determined  in  the  usual  way  of 
evaluating  the  constants  of  a  Fourier  Series,  that  is,  by  multiply- 
ing both  sides  of  the  equation  by  sin 7  and  integrating 

2.  I 

between  the  limits  0  and  2  I.*  We  shall  then  get  for  the  value  of 
the  constant 


Introducing  the  value  of  Bn  in  equation  (78)  we  finally  get 
,       4#o4  /C       .  (         ITS   .  ,1        STTS   . 

/  =  • —  V  T  e      ) cos  ?n  sm  uit  +  o  cos  TTT  sm  3  wit 

7T         T     _/v  T  Zi  I,  A  Zi  I, 


L 

1  57TS     . 

-  cos  -^y  sin  5  coif  +  - 

4^/0  »   \     •       7TS  .1      .      ^  .1  u 

-~nt  '  sin  JT-J  cos  wit  +  -  sm  -^-r  cos 
2Z  3        2/ 


i 


(85) 


where  «i  =  2  TT/I. 

LINE  GROUNDED  AT  BOTH  ENDS 

Equations  (76),  being  perfectly  general,  also  apply  in  this  case, 
but  the  constants  yn  and  /3n  are  different  from  those  discussed  in 
the  preceding  section. 

*  See  Byerley,  Fourier  Series. 


240  FORMULAE  AND  TABLES  FOR  THE 

We  have  in  this  case  the  potential  zero  at  both  ends  of  the  line, 
that  is,  E  =  0  for  s  =  0,  and  E  =  0  for  s  =  I.  To  satisfy  these 
conditions  we  must  have  Qn  =  0  and  ynl  will  have  to  be  a  multiple 
of  TT,  that  is 

7nZ  =  mr,     yn=^>  (86) 

and  0n  =       ^—.  =     .  -  ?  neglecting  a2  compared  with 
IVLC 


The  fundamental  frequency  of  the  oscillations  is  given  by 

/i  =        I—-  (87) 

Equations  (76)  may  now  be  written  for  this  case  in  the  following 
form: 

/—  n  =  oo 

E  =  V  77  €-«<  V  Dn  sin  ^sin  (Zirnfit  -  ^n 

*      C/  v 

J-Sn  COS       ^   '  COS  ( ^  7T71/16  —  Yn)» 


n=  1 

The  fundamental  oscillation  of  the  current  has  its  maximum 
values  at  either  end  of  the  line  but  in  opposite  directions;  and  it 
has  zero  value  at  the  middle  of  the  line.  The  fundamental  oscil- 
lation of  the  voltage  has  its  maximum  value  at  the  middle  of  the 
line  and  is  zero  at  both  ends  of  the  line.  The  fundamental  wave 
length  is  half  the  length  of  the  line.  The  oscillations  contain  all 
the  harmonics  even  and  odd. 

LINE  OPEN  AT  BOTH    ENDS 

In  this  case  we  also  make  use  of  equations  (76)  as  the  general 
solution,  but  the  terminal  conditions  for  this  problem  are 

1  =  0    when  s  =  0  and  s  =  I. 
These  conditions  are  satisfied  by  equations  (76)  if  we  put 

tVJT 

Pn  =  0   and   ynl  =  mr  or  yn  =       •  (89) 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    241 
The  value  of  /3n  is  the  same  as  in  the  previous  case, 


rnr 


mr 


i  VLC    VLOCO 

V,  ^ 

The  fundamental  frequency  of  the  oscillation  is 


Introducing  the  above  values,  equations  (76)  assumes  the  form : 


Dn  gin        Cos  (27m/!*  -  0n), 


E=  - 


cos 


(90) 


For  the  fundamental  frequency  the  current  is  zero  at  both  ends 
of  the  line  and  has  its  maximum  value  at  the  middle  of  the  line. 
The  potential  has  its  maximum  value  at  either  end  of  the  line 
but  of  opposite  signs,  and  its  value  is  zero  at  the  middle  of  the 
line.  The  fundamental  wave  length  is  equal  to  half  the  length  of 
the  line.  The  oscillations  on  the  line  is  of  a  complex  form  con- 
taining a  fundamental  wave  and  all  of  its  harmonics,  even  as  well 
as  the  odd  ones. 


CIRCUIT   CLOSED   UPON  ITSELF 

The  terminal  conditions  for  this  case  are  70  =  Ii  and  EQ  —  EI, 
that  is,  the  current  and  potential  have  the  same  values  for  the 
points  s  =  0  and  s  =  L 

Both  these  conditions  are  satisfied  if  we  put  in  equations  (76), 
which  is  the  general  solution, 


and  hence, 


ynl  =  2tt7r   and  yn  =  — j—  > 
2mr 


i  VLC    VLQCO 


(91) 


The  fundamental  frequency  of  the  oscillations  is 
/.        /5i  1 


(92) 


242 


FORMULAE  AND  TABLES  FOR  THE 


Equations   (76),   in  their  application  to   this   case,  take  the 
forms 

n  =  oo 


2  wns   ,    ~     ,    2  TTHS  ) 

—    --- 


™       4 


«=  1 

\  An  cos  2  7ra/i£  +  Bn  sin 
x^0  (  n     -    27ms   . 

2     ]^nSin-y- 
n  =  1    ' 


(93) 


|  An  sin  2  irnfit  —  Bn  cos  2  irnfit  \ ; 
or  we  may  write  the  above  equations  in  the  following  form : 


V  D 


cos  (2  7m/!*  - 


Fn  sin  —    -  cos 


0n) 


n=  1 

2irns   .    /0       -. 
cos  — j —  sm  (2  7m/i£  — 


(94) 


For  the  fundamental  frequency  of  the  oscillations  equations  (94) 
reduce  to 


cos       -  cos       - 


»y  gT^j  A  sin-^sin^- 


(95) 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    243 


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244 


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CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    245 


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246 


FORMULAE  AND  TABLES  FOR  THE 


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CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    247 


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CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    249 


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CO  CO  CO  CO  T»H 

ddodd 


ooooo 


d  o'  d  o'  d  d  d  o  o'  d  d  o'  o'  d  d  odd  o'  d  d  o'  d  o'  d 


iliil 

ddodd 


ddodd 


d  d  d  d  d 


d  o'  d  d  d 


ss 


88S88  SSSSS  S 

do  odd  dddo'd  ddo'o'd  do'o'o'd  o'dddd 


254 


FORMULAE  AND  TABLES  FOR  THE 


'•S    g 


ooooo  ooooo  ooooo  ooooo 


S 

o  o  d  d  d     *i° 


d  d  o'  d  o 

CSCOCOOCN 

d  d  d  d  d 


CO  CM  i— I  O  OO  CO  T}<  CM  O  CO 

d  d  o  d  d  d  d  d  d  d 

»o  oo  .-H -*  i-,  O5i-icotoco 

d  d  d  d  d  d  d  d  d  d 


ooooo 


OTt<  OOCN  10 

O)  CO  Tt<  CO  I-~ 

dooo  d 


CM  i— i  o  o  o      & 
-*  ^  ^  ^  „      c, 

d  d  d  d  o     o'  d  o'  d  d     d 


<N  CO  CO  CO  CO 

d  d  d  d  d 


lll 


cScO^COCO;         §§5|5SS? 

do'doo     dddod 


Tfi  -^  lOCD  t-- 

d  d  d  d  d 


t^»  to  T*  CM  f— »    O  OO  CO  to  CO 
OOOOO    00  OO  00  OO  OO 

o'  d  o'  d  d     o'  d  d  d  o" 


i-H  OO  to  CM  O  CO 

00  00  OC  O  T-I 

TJ<  •<!<  ^  to  1O  to 

o'  d  d  d  o  o 

O  <M  CO  -*  IO  T*< 

co  i— i  o  r—  »o  co 

ooooo  d 


to  h—  O  *— I  CO 
to  to  to  CO  CO 

d  d  d  d  o' 

00  CM  COO"* 
to  to  to  to  T!< 

d  d  d  d  d 


ooooo     ooooo 


M<O  tOOtO 

CO  to  CO  OO  O 
00  00  00  00  OC 

d  d  d  d  d 


ooooo     ooooo 


Tt<  TJJ  CO  CO  CO 

d  d  d  d  d 


d  d  d  d  d 


ooooo     o 


d  d  d  d  d 

I-H  i-<  d  d  d 


CO  CO  CO  CO  CO 

ooooo 


d  d  d  d  d 
d  d 


T-  OO  to  CM  O5 
COCON-OOOO 

Tf-^^TJ*^ 
d  o'  o'  d  o' 
COM<»OCO»^ 

CM  O  OO  CO  -*f< 

t^t^cceoco 


ooooo     ooooo     o 


10  r-<=5^-ico 

to  to  to  CO  CO 

d  d  d  d  d 

CO^iOOCO 
o  o  oo  r-~  i>- 

d  o'  d  d  o' 


CO  to  t^-  Oi 

r-t-t^t- 


ooooo 
coco  to  T?  S 
o'  o'  o'  d  o'  d  d  d  d  d 


ooooo 

COCOOOOCM 


ooooo 


o  d  d  o  d 


O  *— i  OJ  Tt<  to         CO 

d  d  d  d  o     o' 


d  d  d  d  d 


ooooo  ooooo  ooooo 
d  d  d  d  d 


i-HOOOO       OOOOO 


d  d  d  d  d 
Qf5«o>c5 


o 


5_T?T?: 

o'  d  d  d  o' 


M<  COOOO5O 
J^  CO  CO  S  S. 

d  d  d  d  d 


to  to  to  Tf<  CO 

Tt<  CO  00  O  CM 

d  d  d  d  o 


o  o 


ooooo     ooooo 
o'  d  d  d  o'     o'  d  d  o'  o' 


d  o  d  d  o 


y—  i  to  OO  i—  < 
OO  ^H  CO 
0005050 


ooooo     o 


CO  CO  CO  CO  CO         CO  CO  CO  CO  CO 

o  d  d  d  d  odd  o'  d 


o  o 

llslS 


i|lii  iss&l 

d  d  d  o'  d  d  d  d  d  d 


ooooo  ooooo  ooooo 


ooooo     ooooo 

COCOOCMIO         t-O»-HCOT« 

o'  d  d  d  d     d  d  d  d  d 


do' odd     do' odd     ddddd 


T*  •*}<  T»<  T><  T}<         CO  CO  CO  CO  CO 

ddddd     do' odd 


d  o'  d  o  d     o  o  o  o  d 

o'  d  o'  o'  o'     odd  o  o' 


i^gii 

n  co  co  co  co 
d  o'  d  d  d     d  o'  d  o'  d 


S'o      o^IcN 

)  T*  Tfl  T(H  Tt< 


CO  COCO  ( 

o'  d  d  o'  d     o'  d  o'  d  d 


ooooo  ooooo  ooooo 


,  |S    SSScSg 

d  d  d  o  d      o  o  o  d  d 


OCXI  •*  COOO 

ddddd     do odd 


OOOOO        OOOOO       1-1 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    255 


I   + 

t"^  J^x 


88388    SS3SS2 

dodo'd  dddo'd  do' odd  do'dod  ddddd 


do' odd  ddddd  ddddd  do  odd  do  odd 

•*t<  Tt<  CO  OQ  i— I  Oi  !>•  IO  CO  i— t  OS  O  CO  OS  1C  O  IO  O  »O  O  1C  O  CO  !>•  O 

COCOCOCOCO  CSC^C<IC<IC<I  ^-Hi-Hi-HOO  OC5O5OOOO  t^-COOiO»O 

d  o'  d  d  d  d  o  o  d  d  ddddd  d  d  d  d  d  o'  d  d  d  d 


do'ddd  ddddd  o'ddo     oo     oo 


TO  CO  CO         COC 
IM(MC^  (M( 


ooooo  ooooo  ooooo  ooooo  ooooo 


ooooo  ooooo  ooooo  ooooo  ooooo 


83858    5883*3    3:2222    3888$    SSS3S 

do'o'dd     o'o'ddd     ddddd     o'dddo'     do'o'dd 


o'  d  o  o'  o 


ooooo     ooooo 


ooooo     ooooo 


>  10  o  kc     o  10  os  co  r>- 

)-^^fCO         CO  O4  I-H  i— i  O 

>  CD  CD  CD         CD  CD  CD  CD  CD 


OOOOO    OOOOO   OOOOO    OOOOO    OOOOO 


ooooo 

<N  <M  (N  ?5  CNI 


O  O  O  i-H  i-H          T-l  rH  i-H  rt  i-( 

ddddd     o'  d  o'  o'  d 


O»-IC^-^l>O         COt^CDO'H 

do'ddd     do  odd 


Oi  OO  t-- 

OOO 


§^^So5 
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o'do'dd 


o  o  o     o 


S^coos^  coSoooco 

rH  i-H  ^H  r-t  C<I  CM_<M(>JCOCO 

ddddd  o'do'dd 

d  d  o'  o  o'  d  o'  o  d  d 


oo  oodo'd 

o'  o'  d  d  d     d  d  d  d  d 


ddddd 


o 


ooooo     ooooo' 


§C<I  ^  l>-  Oi         i—  '  •**<  CD  OO  i—  (         CO  »O  OO  O  C>l 
0000         rt^^H^-lC  Ca(M<MCOCO 


o  o  oo  od  do' 


s«i§§  si  ii  Jo  Si! 

o'ooo'd     ddddd 


ooo  o  o  o 


OOOOO 

ddddd 


OOO^i-H        ^H^H^H^I^H 

ddddd     ddddd 


OO        l^»  t^. 


do'  odd     o'dddo 

<MiCt^05i-<        (M  <M  <M  (N  i-H 

CO  CNI  *—  f  O  O         O  OO  t^-  CO  IO 


do' odd      ddddd      o'do'od      do' odd      o'dddo 


256 


FORMULAE  AND  TABLES  FOR  THE 


1  g 

o    * 


ffl 


+ 


O  C<1  -^  CO  OO        O  C<I  Tj«  CO  OO 
«0«0»OiaiO         cpcOCOcOCO 

o'  d  d  d  d     ddddd 


O  C<1  -^1  CO  OO 
!>•  !>.  S  «>.  ?i 

o  d  d  d  d 


ooooo  o     i- 


<  1C  CO  t"-  OO  Oi  O>  < 
>  *— I  CO  IO  t-—  Oi  *— I  C 
(CO  COCO  CD  cot^l 


ooooo  ooooo 

COCOO5»-IOO    »Ot-OOO5O 


ooooo  ooooo 


ooooo 

IsSss 

o  o  o  d  d 


d  d  d  d  d 


ooooo  ooooo  o 


ooooo  ooooo  ooooo 


ooooo 
1000  ooooo  ooo'o'd 


ooooo     o 

£zJ2t2ttco     o 

d  d  o'  d  d     d 


ooooo  ooooo  ooooo 


d  d  d  d  d 


d  d i  d  i-i  i-i     i-i 


CD  CO  CO  to   *O  to  tO  iO  iO 


ooooo  ooooo  ooooo  ooooo  ooooo  o 


o  o'  d  o'  d  o'  d  o'  d  d  ddddd 


O  i-H  C<l  CO-<* 

d  d  d  d  d 


ooooo  o 


I^H  OO-*    O 


1^  CO  »O  CO  C^         i-H 

ooooo      o 


HOOOO        OOOOO        OOOOO        OOOOO        O 


d  d  d  d  d 


OO  O  *— '  CO  »O 

d  d  d  o'  o' 


ooooo  ooooo  ooooo 


d  d  o  d  d 

^IH^S 

d  d  o'  o'  d 


ioo  °. 

)  CO    CO 


ooooo  ooooo  ooooo  ooooo  ooooo  o 


T-HOOOO        OOOOO        OOOOO        OOOOO        O 


m  10  co  co  CD 
d  o'  d  d  d 


t>»  O  i— '  CO  *O 

d  d  o  d  d 
mt~*.a*T*  co 

-+•  ~-~  oi  01  ^-H 
>C  1C i  10  10  1C 


OOOOO       OOOOO       OOOOO       »-H 


>  Ci  OO        t->- 

SCO  CO         CO 


ooooo  ooooo  ooooo  ooooo  ooooo  o 


>»-i(M         CO- 


cocococo 
do'  odd 


»O  IO  »O  IO  -^ 

C)  CO  Tt<  10  CD 

d  d  d  d  d 


I>»OOOO»— I 
•*  Tt I  -^ I  i«  10 

d  d  odd 


d  d  d  d  d  o 


,_,  ^H  ,-H  ,-<  ,-1   OOOOO   OOOOO   OOOOO   OOOOO   O 


d  d  o'  d  d  d  d  d  d  d 


CD  OO  Ci  i—  <  CO 

t^.  t^.  t^-  oo  co 


b-OJrH  (MCO 


ooooo  ooooo  o 


ooooo  ooooo  ooooo 


^  tO  CD  CD  CD 

d  d  d  d  o' 


c^cogco1      3, 
ddddd     d 


»o  cor^ooo 

CO  CO  CO  CO  CO 

d  d  d  d  d 


do  odd     do  odd     o'o'o'o'd     d 


- »O        CO 
t>-        t^ 


^I^H^^O  ooooo  ooooo  ooooo 


T— I  O>  OO  CO    ^ 

_b-coeoco   to 
d  o'  d  d  d  d 


od     o         o  o  oo  o  o 


8 

o'o'ddd     •-! 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    257 


+ 


CQ 


+ 


8SSSS 

dddo'd 


o  o         oooo 


OOU50 


d  o'  d  d  o  o 


o'ddo'd 


o'd  odd 


d  o'  o'  o  d 
odd  do 


IC003  cOCOOOO 

d  o  o'  d  o'     o  d  d  o'  d 
odd  do     o'o'o'dd. 


o 


ooioor^co 

CM  i—  «  »-*  I-H  i-H 
CO  CO  CO  CO  CO 


co  co  i"*  t*-  "^ 

oo  o  c<i  co  ic 
p  I-H  i-H  i-i  1-1 

o'd  odd 


»-«  oo  ic  »-i 
is»  oo  o  CM 
1-1  T-I  c-j  CM 

o'd  od 
•«»<  co  CM  ic 

C3  OO  OO  t-- 
O4  CM  CM  CM 


us  I-H  r^  co  o>  usi—i^co 
ic  t^  oo  o  I-H  co  ic  co  oo 
CM  CM  CM  co  co  eo  co  co  co  < 


o     o  oooo         o'oo'o 


»•<  CM  eo  •**  10     us  - 

CO  1C  •**<  CO  CM         i—  ' 
N  C^  IM  (N  M         W  I 


d  o'  d  d  d 


o 


COIOKMIM 

COCX3C»OOQO 

OOOOO 


O  T*<  O  1^5  T-^ 

s^^Sci 

ddddd 


oooooooooo 
d  o'  d  o  o 


. 

dddo'd     ooooo 


o'  o'  d  d  o' 


ooooo 


O  i—  t  CO  lO  O 
OOOOO 

ooddd 


coor~coo 

OOOi-HCO-^ 
O  »—  t  i—  I  »—  <  i—  < 

ddddd 


CO  GO  O  T—  'CO 
i-H  i—  (  i—  I  C^l  C^l 

ddo'od 


)  CO  t^  O  CO 
•  IT-.COCDIO 
ICMIMIMO* 


1^-  OO  Oi  O 
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C^<MIM<M 


cocSScoco; 
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8O  OiOO 
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OC^iOl^-O 

p  p  O  O  ^H 

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i-l  .-H  ^  CV4  1  C» 

o'do'do' 


ddddd     ddo' 


CM  CM  CO  CO  CO 

OOOOO 

05  COOO  GO  t-~ 

o'  o  d  d  d 


*-*CO       i—  '  »O  O  Tf  t*» 
COiO         OOOCOlOt^- 


ooooo 

(M  t^CM  t^CM 
t-- CO  CO  1C  1C 

o'  d  d  d  d 


OC^^t^-05 
«O  US  US  «S  KS 

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ooooo 


85SSS 

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CO  1C  t^  OC  O 
CM  CM  CM  CM  CO 

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CM  CM  CM  CM  ^H 


CO  CO  CO  CO  CO 


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dodod 


rtiTjHco(M^ 
oooooooooo 


oo 


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t^.t^t^t>. 


t--  ^HlCO:  CM 

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338co-£ 


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i— i  ^  CO  OO 
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d  o'  o'  d  d 
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ooooo' 

t^-  t^-  t^-  CO  CO 
CM  CM  CM  CM  CM 


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t-^.  O5  i—  <  CM  -^ 
O  p  ^H  T-H  ^H 

oo'ddd 


COIM  t^CJt- 

lCt>-OOO^H 
T-H  ,-H  rt  <N  <M_ 

do'do'o' 


CO-^COt^-Oa 
Cq  <M  <M  (M  <M_ 

ddddd 


1C  «  CO  ^  -* 
OT—  'CO-^CO 
CO  CO  CO  CO  CO_ 

ddddd 


COiCl 
4  (M  IM  ( 


<M         «  I  CM  I  Cq  I  <M  <M         (M  (M  i-H  »-H  i-H         r-(  r-l  -i-l  ft  j-> 


iCO  ICO 
(MiCt^O 
OOOi^ 

ddddd 


O  >COCDO  Tt< 
O  <M>CI>-OCM 
i^  1^1—  li—  »  CN!  C^ 


O5  -*  OOCO 
Tt<  t^  Oi  cq  • 
dOiCNCO 


o 


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t^OSt-H^fCD 
COCO1^^^ 

dddo'd 


O5CM  US  OOO 
OO  i— *  CO  1C  OO 

d  o'  d  o'  d 


ooooo  ooooo  ooooo  ooooo  ooooo 


CO  ^H  CO  ^H  CO 

r-o>o<Mco 

O  O  i—  '  i—  I  i—  t 


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oocoic 

C^COCOCOCO 

dddo'd 


odddd  o'ddo'o'  ddddd  ddddd  o'ddo'd 


258 


FORMULAE  AND  TABLES  FOR  THE 


i    * 


« 


+ 


OOOOO   OOOOO   OOOOO   OOOOO   OOOOO   1-1 


OOOOO   OOOOO   OOOOO   000' 


§t^>  OO  O  O          T-H  i-H  T— 4  i— I 
TjtecciiM      i-iooioo 


ooooo 


§^CO(Mi— I        OOOt^-COiO        COC-1OOOO        O 
OOOO        OiOiOiOlO        10  IO  »0  "*  •$<        Tt< 

ooooo  doodd  o'odod  d 


ddddd 


GO  Oi  *— '  Ol  -^ 


ooo  oo 


dodod  -00000  o'odod  d 


i-iOOOO   OOOOO   OOOOO   O 


SS'oS^ 

ddddd 


i--  r~  t^  oo  oo 

d  d  d  o"  d 


OOOOO   0000.-1 


o'ddo'd  o'ddo'd  o'o'do'd  o'ddo'd  odd  do  d 


OOOOO 
OCOO  CO(M 


OOOOO 


t^o  IM  ^o 
ddddd 


e<i  ^o  osr- 


OOOOO   OOOOO   OOOOO   O 


OOOOtMTH         IOOI>-OOOO         OOOO 
*— <  CO  CD  OO  O         C<l  •**"  O  CO  O         C^-^ 


ooooo 

O  (M-^IOCO 

t^.  o.  o  co  o 

d  d  d  o'  o' 


>OI>-I 
I  •<*<  0( 

o'  o'  d  o  o 


00  t-  < 


SSSS8 

o  o'  d  d  o' 


ooooo 

OSOOOt^-CO 

S i  o  §  S  «5 
d  o  d  d  d 


co  »-<  oo  10  »—  « 

C<l  -rfi  10  t^-  O 
Oi  O  O  O>  O> 

o  d  o  o'  o 


IO  ifl  i  1O  IO  >O 

o  o'  o'  o  d 


giro's  j§ 

»o  -^  ^  •*»<  Tt<      *# 

dodod     o' 


Jo^2§S^      Sooooio      caco^ioco      oo 


i— 4  C^  CO  CO 

SO-H  cj 
000 


O  O  00  O   OOOOO   OOOOO   OOOOO   OOOOO   O 


sssss 

d  d  d  d  d 


d  d  o'  d  d     d  d  o'  d  d     d 


d  d  d  d  o' 


sgsgg  sssss  s 

ddddd  d  I-H'  »-<  -i  i-i  «-! 


ooooo  ooooo  ooooo 


•^t4  co  c^  »— »  o> 

1O  1O  »O  IO  •>* 

d  o'  d  o"  d 


f->oe<i  os  eo  co 

oo  r-  o  ^  co  ci 

-*  •*  •* i  it i  -^  -* 

ooooo  o 


IIOOO  -r-iT^t^OC^  Tt<  O  OO  O  i— i  O1C 
it— i  <N  -^*/tiOOOOO5  O  i— "  C^J  -^  »O  COt 
i  ^f,  ^fl  ^^^^4^  iO  IO  IO  IO  IO  IO  * 


CO  O  CD  O  c 


ooooo  ooooo  ooooo  ooooo  ooooo  o 


,_, ,_,  ,M  ^H  T-H   I-H  1-1  I-H  T-I  O   OOOOO   OOOOO   OOOOO   O 


§••*  ot^  oo 
o  S i  o  o 
d  d  d  o  o 

CO  »O  »O  Tf  CO 

ooooo 
d  o'  d  d  o' 


1III1  Isiii  ilsli  1 


ooooo  ooooo 


ooooo  ooooo 


ddd 


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IO  »O  IO  Tf I  rjt 

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o  o  t-- 


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ddddd 


O(M  T»<  OOO 

o  o'  o  d  o 


§SiSS§8    SSSSSS    8 

d  o  o  o  d      ooooo      i-i 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    259 


PQ 


8§S§§  22222 

odd  do  do'  odd 


o    o     o    ooo 


OOOl-^tfSCO  i-H  O  00  »O  CO 

ddddd  dodod 

OO  OO  l*~  CO  *O  CO  i— '  OO  »C  CM 

OOOOO  O  O  OO  OO  OO 

OS  OS  O  O  03  OS  OS  OS  OS  OS 

d  d  o'  o'  o'  o'  d  o'  o'  o' 


SiSal?    3SS8S5    SEoSSS 
d  o'  o'  d  o'     ddddd     d  d  d  d  d 

o'  o  d  d  o'     o'  d  o  d  d     o'  o'  d  o  o 


OSOOOOt^- 
O  >->  CO  >0 
1-H(M<M<M( 


ooooo     ooooo 
COCOCN^HO 


OOOOO   OOOOO   OOOOO 


<-*coco 


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t>-  O  CO  lO  OO 

w  co  co  co  co 
d  d  d  d  d 


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»o  10  10  o  «o 


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IIM-HOO 


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O  O  O  O        O  ^  r-l  12  T-H 


o  o  o  o 


co  *-H  t—  (  o  o 

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doddd 


~ 

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ooooo     ooooo 


is 


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d  d  d  d  d 


d  o'  o'  d  o 


o'  o'  odd     o'  d  d  o'  d 
d  o  o  o'  d     o'  o'  o'  d  o' 


IT-H  ooiO 
)O^H  CO 


ooooo     ooooo 
cococococo      eococococo 


ooooo 

00  <M  COOS  T* 
CO  CO  CO  CO  CO 


o  ooo 


_ 

odd  do 

ifliOTfCOlM 

o'  o'  d  d  d 


COCOOr-<^ 
i-l  i-H  T-H(M  CSI 

odd  do 


o'  d  d  o'  d 


OOCOO'O  *—  *  t^>  CO  Ot  -^  OO-^OOC^t^ 

t^OCM^l^  OfMOt^O  SllfttSSS 

(M  (N  CO  CO  CO  -*  IT*  |  •*  I  ^  I  >O  »O  «0  "O  «O  CO 

do'do'd  do'  odd  do'do'd 

«OCOO>OO  "*OCOt^O  COCOOOOIM 

o'  o  d  d  d  d  o'  o  o  o'  o'  d  d  o'  d 


O  0000 

o'  d  d  d  o 


ooooo 


^o       -<*t 


ooooo 

CO  CO  CO  CO  CO 


CM  CM  CO  CO  CO         COCOCO-«*Tt< 

o  o  o'  d  d     o'  d  o'  d  d 


3§!s§S 

ooooo     do' odd 


>  t^  CO  O  l^-  -^  O 
><M»OOOO  COCO 
)OOOrH  T-Hr-l 


CO  CN  OO  -^f  < 
COOS  i-H  -5<l 
CM  CM  CO  COt 


OOOOO   OOOOO   OOOOO 


ddddd  o'dodd  do' odd 


•^•^COCOC^         t— *  T— i  O  O  OO 
00  00  00  00  00        COttCCt^t*- 

d  o'  d  o  d     d  o'  d  o'  o' 


O  i— I  CO  IO  t^- 

o  o  o  o  o 

o'  d  o'  d  d 


rH  i-l  i-t         i-l  i-l  <M  <M  CM 


§iiii  i 


iS§ 

o  o  o'  o'  d 


i-IOO-^OCO 
COCOCOCO 


I  O  OCOC 


CO  CM  OOTft  O 
CMCMCMCOCO          COCOCOCOT 

o'dodd     ddddd 


)^CNICM         CM^HOOOCO 
•COlO-^         CO  CM  »— I  OS  OO 


ScM-^COOO         OcM-^COOO         OcM-^COOO         O  CM  ^*<  CO  OO 
Hrt^H^-l^H         CMCMCMcMcM         COCOCOCOCO         SjS  ^  ^1  -*  •* 

dodod  o'dodd  odddo  dddo'd  ddddd 


260 


FORMULAE  AND  TABLES  FOR  THE 


B 


S       DQ 

«       § 

°       ^ 

T'? 


M        „ 

^      § 


O  C<l -*tt  CO  OO 

cococococo 


OOOOO   OOOOO   OOOOO   OOOOO   OOOOO   »-i 


COCO  t-- i-l  ifi 
f-~  O  CN  IO  1^- 
CO  t*- t>- t"-  F— 

d  d  d  d  d 

CO  CO  1O  1C  TfH 

d  d  o'  d  d 


o>  c^  -*t«  co  oo      i— i  co  *o  t^  Ci 
o*  o  o  o  o     o  o  o  o  o 


O          COr-tCO-H< 


ooooo  ooooo  ooooo 


CO  CD  CO  O  Is- 
^  Tj<  fcO  *O  iO 

o  o"  o  o  o 


>COOC<M         COO  COI 
}t-~OO<S        1-1  CO-*  I 


ooooo  ooooo  ooooo 


o'  o  o  o  o' 


o'o'ddd  d 

T-IO5COCOOS        >O 

o  d  o'  d  d     o' 


o  o  o  o  o 


oooooooooo 
d  o  d  d  o 


OOOCOIOS.        05^-i-^lCDOO 
t>.00(»0000        00  05  0>  05  05 

o  d  o  o  o     d  o'  d  d  o 


o         o         oo 


. 

SScoiot-:     C»OC^TJHIO     K 
OOOO        C^'"<»~<'-^'Mi        I-H 


t-,  o  •<*<  co  ^H 

COO  CO  «O  «O 

o'  d  d  d  d 


ooooo     o 


33J338 

o'  o'  d  d  o 


o  d  d  o  d  d  d  d  d  d 

IO  CO  *— <  O  OO  CO  -ft<  CO  *— 'OS 

^-l^i-l^-iO  OOOOOS 

,_,'  ^H'  ^H'  ,_'  ^H'  ^  ^-i  _('  i-!  O 


OOOOO       OOOOO        O 


COCOCOIMO 
t^-iOCO^—  <Oi 
O  O5  Oi  C5  OO 

d  d  d  d  d 


OO  OO  OO  OO  t-—        l^» 

d  o'  d  d  d     d 


t^.00000000 
C<1*^OO1OO 

o  d  d  o'  d 


lOl^OS^H  CO        «• 

ddddd     ooooo 


o 


!2S3    S 


ss 

d  o'  d  o'  d 


ooooo      o 


OOOOO 


CO  -^  CO  I^  OS 

d  d  d  d  d 


SSSSos 

^H'  ^-i  ^|  CD  O 


ooooo 

Ci  OO  t^-  CO 


d  d  d  o'  d      o 


ooooo      ooooo      o 


o 


o 


1O  C5  O<M  CO 
CO  OO  »— '  CO  kO 
1>-  t>-  OO  OO  OO 

d  d  o  d  o" 

iO»O>O  ^  <M 
»O  Tjl  CO  <M  i— » 

t-- •  t^  c~  t^  t>- 

o'  d  d  o'  d 


r-(  OO  Tt<  O  kO        O 

eo  t-  g  m  c^   ^ 


o  So  i-  •»  S 

t-  co  CO  CO  CO 

d  d  d  d  d 


d  o'  d  d  d 


§  10  S  C^l  O         O 

d  o  d  d  d     d 


SSSSS55 

o'dddd 


OOOOO 


§o  'C^: 
coco 

id 

)  CO 

;g 


OOOOO 


100 


o     o  o  o 

CO-*  O>  t^US  CO 

Tti  CN  OS  t^  10  CO 

o'  d  o'  d  d  d 


o 

0>0»-i<MCO 
t^-  t>-  CO  kC  *<tl 

o'o'ood 


ooooo     ooooo 


COCN^OOS 

r*  r—  i-*  t>-  co 

ddddd 


co  to  ^  co 
co  co  coco 

d  d  o'  d  o 


d  d  d  o  o 


<M  00-^<  O  >O        O 

d  d  o  d  d      d 


i— t  co  i— 1 10  o 

gss?ss 

d  d  d  d  d 

S'   Oi  CC  Is*  O 
OO  (^  >O  g 


l>-  OO  O5  »- '  Ol 
1C  «O  IO  CO  CO 

d  d  d  d  d 


ooooo 
1  d  d  d     o"  o  d  d  d 


ooooo     o 

OO  OO  l>- !>•  l^-        t*- 

ddddd      o 


ooooo  ooooo  ooooo  ooooo 


o  M  **  co  oo      o 

OSO  03  05  OS          O 

d  d  d  o'  d      ^ 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    261 


^3 
*•» 


e     OQ 


o    < 

T5 


H     « 


8§SSS    2S3SJSS    8S3SS85 

o'  d  o  d  d     d  o'  o'  o"  d     d  d  o  d  o 


o  d  o  o  o     o         o 


p  p  p  p  TH  TH  TH  C^  <M  <M  CO  CO  CO  CO  -* 

o'  d  d  d  d  d  d  d  d  o'  o'  d  d  d  d 

US  •>*•«*<  CO  <M  O5  t~  •**<  TH  t^-  Wt^CMCOO 

Ttl  ^  ^  •<**•*  COCOCOCOIM  CMTHTHOO 


d  o'  d  d  o'  o'  d  o'  d  d 


ddddd  d  d  o'  d  o'  o'  d  d  o'  o' 


O5  i-l  CO  IC«O  CO  t~  OO  C5  O5 

CO  SO  OOO<M  -*COOOO<M 

codfc  CO  •*  -^  **_  -^  •*  tf>  "5 

o  o  o  o  o  o  o  o  o  o 


O        O  O  O 


»O        CO  C^l 


o     o  oo  o 


o'o'ddd     o'dddd 


oo         oo  oooo 


CO  CO  CO  CO  •*         >*  •*  •*  •*  >O 

odo'dd     o'o'do'd 


I-H  i—  I  O        OO  CO  CO  O  Is- 

^§^      cococococo 


o'  d  d  d  o'  o'  o'  d  o'  d  ddddd 


ooooo  ooooo 

TH'  TH  TH  TH  TH     O  O  O  O  O 


^  CO  (M         C<I 

ocoS      ^ 


ooooo  ooooo  ooooo 


ooooo  ooooo 

-<1<  •*  CO  CO  CO         COCOCOOOOO 


O  O  O  O  T-I         i-l  i-H  (N  <M  CM         CM_  CO  CO  CO  -^ 

o'do'o'd     o'dddd     o'do'dd 


co  »o  oo  *-*  -rf*      co  os  i— i  ^t<  r>- 

^  Tt<  -*  1C  U5        US  U5  CO  CO  CO 

d  o'  o'  d  o'     d  d  o'  o'  d 


n  o'  o  o'     do' o'  o' o' 


ooooo     ooooo     ooooo 

3n$S38S    SSSSg    S22§§ 


ooooo  ooooo 


O  O  rH         rH  rt  <M 


CO  CO 


•*  •*  ••*  >O  SI        U5  U5  CO  < 


ooooo  ooooo  ooooo  ooooo  ooooo 


O  O  O  O  O          O  O  O  O  O 


ssssa  mm 

dddod  ooooo  o'dooo' 


o  o  o  o 


S(M-*COOO        OIMr»<OOO        OlMTflcOOO 
0000         THTHTHTHTH         <M<MC<I<MC^ 

doodd     o'o'ddd     doood 


oooo     oo 


262 


FORMULAE  AND  TABLES  FOR  THE 


I    § 

<S  «. 

1°. 

X 


% 


o      OCM-*COGO      OIM^COOO      O<MT*COGO      o 

O         t^t^t>-r^l>.         Op  00  GO  OO  OO         O5O5O5O5O5         O 

ooooo     ooooo     o'  d  o'  d  o'     d  d  d  o  d     o  o  o'  o'  d     »H 


S££§8    £§88gg    SSSSS    §322 

do' odd     do  odd     d  i-i  i-l  ,-i .-!     ,-i  ,-i  ,-i  ,-i 

"5-^(NOOO          »«<MO5IOi-l          COrHCOO-*          OO^H^t^. 


i-iOOOO   OOOOO   OOOOO   OOOOO   OOOOO   O 


>  »O  CM  O5  IO        T-H  t-~  CM  t-*- CM         t*>-  t-H  ^  I 

)  »O  t^  GO  O         CM  CO  »O  CO  GO         O5  »— t  OJ  ( 


ooooo 

CO  CO  CO  CM  CN 


ooooo     ooooo 


5oooooo< 
o'  d  o'  o'  d 

OOOCOCOO 
T-|  ,-H'  ^i  d  O 


o 

COCNl  t*-  <Mf-  T-I 

^f  <M  O  t^-  ^t1  Ol 

O  O  CO  OO  GO  GO 

o  o  d  d  d  d 


Tf<  CO  O5  *— I  T^ 
OOOOOOC5O5 


t^-  t--  !>.  t^.  C 

do'o'do     ooooo     001 


<MO 
1^  O 


o'o'o'o'd     o'o'o'o'd     do  odd     do' odd     do  odd     d 


d  d  o'  d  d 


ooooo 


ooooo     o 


<oo     ooooo     o 


ooooo     ooooo 


SCM-*  COt^. 
(--  05<-H  CO 
O505O5O  O 

d  d  d  i-I  »-I 


5  r*  o  I-H  c 


d  d  d  d  d 


Oi  OO  l^-  iO  -^ 

o'  d  o'  o'  d 


co »— '  o  oo  r— 
d  o'  d  d  d 


tOO-^  00  (M 
lJ5j*C^Og 

d  d  o'  o'  d 


1^*  O1 
-  O5  »— < 
»O  <o 


ooooo     ooooo 


oo  05  o  o  o 

COOt-lO 


OgfOCOO  CM 

S  OO  OO  00  O5  OS 

d  o'  d  d  d  o' 

<  d  d  o'     o'  d  d  d  d  d 


§O1  T}H  t>- O5         C^l  •**<  CO  O5  i-H         CO  1C  OO  O  CM         Tt*  O  OO  O  CM         CO  »O  t^- O5  O         01 
t^t»l>.t^.        OOOOOOOOOJ        O5OSO5OO        OOOrH^-l        1-1  r-n-H  ^-i  (M        CM 


OOOOO       OOOOO       OOOt 


•^  C5  t^  -fi  ^ 

r—  10  ^  co  c^ 

OO  00  GOOD  GO 


s  GO  CO  »O 


OO5CO        COO3(M»COO        O 
?*  CO  CO         CO  CO  CO  CO  1C         1C 

o'  o'  d  o'  o'     o'  d  d  d  o'     d  o'  o'  o'  o'     o'  d  o'  do'     o  o  o'  o'  d     d 


III     I 


ooooo 

CO  CM  r~  CM  CO 


ooooo 


ooooo  ooooo  ooooo  o 


sssss 

>-!  d  d  d  d 


ooooo  o 


o'do'dd  ddo'dd  do'o'o'i-* 

0300000000        COOOOOOoS        ?2t^t2^?2 

o"  d  d  o'  d  d  d  o  o'  d  o'  o'  o  o  d 


SoOO         §5^10 


!S8§2 


o'  o'  o  o  o 


d  o  d  d  d     d 


000*00  ooooo  oo'o'oo  ooooo 

CQ^O}'*&O*  CO  O  O5  C^J  ^  CDOOOiCiOS  OO  t—  <O  i^  CO 

lO^C^^HOi  OOCO-^CO^H  O  t^-  lO  CO  i— *  OSt^-^OCO^H 

C^(MCSC^T-H  i-«  I-H  »-(i-ii-l  OOOOO  OSOiOiOO 


O  i— '  CNJ  Tf  1O  CO 

OOGO  OOOOOO  00 

d  d  d  d  d  d 

i-H  OOlOtM  GO  •** 

05  co  -*t<  CM  O5  IS- 
C' d  d  d  d  d 


o  o 


8S835S3    8 

d  d  d  d  d      T-«' 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    263 

REFERENCES 

C.  P.  Steinmetz,  "Theory  and  Calculation  of  Transient  Electric  Phenom- 
ena and  Oscillations,"  section  3,  Chapters  II  and  III. 

G.  Roessler,  "Die  Fernleitung  von  Wechselstrome."    * 

J.  A.  Fleming,  "The  Propagation  of  Electric  Currents  in  Telephone  and 
Telegraph  Conductors." 

C.  P.  Steinmetz,  "Elementary  Lectures  on  Electric  Discharges,  Waves 
and  Impulses  and  Other  Transients." 

M.  I.  Pupin,  "  Propagation  of  Long  Electrical  Waves,"  Proc.  Am.  Inst.  E.  E.} 
Vol.  16,  p.  93,  March,  1899. 

A.  E.  Kennelly,  "The  Process  of  Building  up  the  Voltage  and  Current  in 
Long  Alternating  Current  Circuit,"  Proceedings  of  the  American  Academy  of 
Arts  and  Sci.,  Vol.  42,  p.  701,  1907. 

A.  E.  Kennelly,  "  The  Equivalent  Circuits  of  Composite  Lines  in  the  Steady 
State,"  Proc.  Am.  Acad.  of  Arts  and  Sci.,  Vol.  45,  November,  1909. 

Discussion  on  Output  and  Regulation  in  Long  Distance  Lines  and  Calcu- 
lation of  the  High  Tension  Line,  Trans.  Am.  Inst.  E.  E.,  June,  1909. 

A.  Blondel,  "Long-distance  Transmission  Lines,"  Eel.  Elect.,  Vol.  49,  pp. 
121-129,  161-166,  241-249  and  321-333,  1906. 


264  FORMULAE  AND  TABLES  FOR  THE 


CHAPTER   VII 

MATHEMATICAL    FORMULAE 

EXPONENTIAL  AND  LOGARITHMIC  FORMULAE 


/y»  2  /J/V»  3  /y  4 

•*/  ./*^/          I       *k 

~~  ol        o~f    '   Tf 

At  O  1  ~r  I 

=  cos  x  +  j  sin  x. 

e-;x    _    cog  ^   _  j  gjn  ^.^ 

c/x  _|_  e-jx  —  2  cos  x. 
e'x  —  t~ix  =  2  j  sin  £. 
ex  =  cosh  x  +  sinh  #. 
e~z  =  cosh  a:  —  sinh  x. 
ax+y  =  axav. 

ax 

ax-y  =  axa-y  =  _. 

a? 

If    x  =  au,  then  u  =  loga  x. 
If    x  =  eu,  then  w  =  logeo;. 

=  log  x  +  log  y. 
W 


log  (xn)  =  n  log  a;. 


logic  ^  =  loge  X  logio  e. 
log€  a;  =  logio  x  X  loge  10. 
logio  e  =  0.434294. 
Iog6 10  =  2.302585. 
logio  €  X  log.  10  =  1. 
e  =  2.718282. 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    265 

l/x-  1\5  . 


TRIGONOMETRIC  FORMULA 

sin  (x  ±  2/)  =  sin  #  cos  ?/  ±  cos  x  sin  £/. 
cos  (x  ±y)  =  cos  x  cos  y  =F  sin  x  sin  t/. 

i          ,        tan  x  =1=  tan  y 
tan  (x  ±  y)  =  —  —  —  • 

1  =F  tan  x  tan  y 

f          v       ctn  x  ctn     T  1 
ctn  x  ±  i 


ctn  x  ±  ctn  ?/ 
sin  x  -b  sin  T/  =  2  sin  J  (x  ±  y)  cos  J  (x  =p  y). 
cos  z  +  cos  y  =  2  cos  J  (x  +  £/)  cos  i  (x  —  y). 
cos  a:  —  cos  y  =  —  2  sin  |  (#  +  y)  sin  £  (x  —  T/ 

sin  x  ± 
tan  x  ±  tan  ?/  = 


ctn  x  db  ctn  v 


cos  x  •  cos  y 

sin  (x  ±  ?/) 
- 


sin  x  •  sin  y 

x3      x5      x7 

-  +  ---+  . 

/>«2         /v«4         /v»6 


sin  x  =  —  \j  (exj  —  e~X1'). 
cosx  =      e 


A  cos  x  ±  5  sin  a;  — 
where         tan  ^  =  -r 


cos 


A$mx±Bcosx  =  VA2  +  B2sin  (x  ±  0), 

T> 

where      tan  0  =  -j  • 
^1 

sin2  z  -  sin2  ?/  =  sin  (x  +  y)  sin  (x  —  y). 

cos2  x  —  cos2  y  =  —  sin  (x  +  z/)  sin  (x  —  y). 

cos2  a:  —  sin2  ?/  =  cos  (x  +  T/)  cos  (x  —  y). 

cos2  a;  —  sin2  x  =  cos  2  z  =  1  —  2  sin2  x  =  2  cos2  x  —  1, 

sin  2  z  =  2  sin  x  cos  x. 

(cos  z  db  j  sin  z)n  =  cos  nx  d=  j  sin  wz. 


266 


FORMULAE  AND  TABLES  FOR  THE 


—  i. 

90°  ±  z. 

180°  ±  x. 

270°  ±  z. 

360°  ±  z. 

sin   .  . 

—  sin  x 

-f-cosx 

•T~sin  x 

—  cos  x 

isin  x 

cos         .  .         ... 

-j-cosx 

-Fsinx 

—  cosx 

-f-sin  X 

-(-(jOS  X 

tan  

—  tanx 

-Fctnx 

rttanx 

-T-ctna; 

rttanx 

ctn  

—  ctnx 

-Ftanx 

-l-ctn  x 

-Ftanx 

-l-ctn  x 

sec  

-{-sec  x 

-Fcscx 

—  sec  x 

-4-P.SP,  X 

-j-sec  x 

cose 

—  esc  x 

-j-sec  x 

-Fcsc  x 

—  sec  x 

irsft  x 

HYPERBOLIC  FORMULA 

sinh  (x  +  y)  —  sinh  x  cosh  y  +  cosh  x  sinh  y. 
sinh  (x  —  y)  =  sinh  #  cosh  ?/  —  cosh  x  sinh  y. 
cosh  (a?  +  y)  =  cosh  z  cosh  y  +  sinh  z  sinh  y. 
cosh  (x  —  y)  =  cosh  x  cosh  ?/  —  sinh  #  sinh  y. 

tanh  #  +  tanh  y 
farcMs  +  a-^tanhstonhy- 


,  ,          N       tanh  a:  —  tanh  y 
tanh  ($  —  #)  =  -—     — r—    — r^- 
1  —  tanh  x  tanh  y 

cosh2  a;  —  sinh2  a;  =  1. 
1  —  tanh2  x  =  sech2  x. 
1  —  ctnh2x  =  csch2a;. 

A  sinh  x  ±B  cosh  x  =  VA2  —  B2  sinh  (a;  ±  <j>) 

n 

where  tanh  <J>  =  -j  • 

A  cosh  x  d=  B  sinh  x  =  VA2  -  B2  cosh  (x  ±  0; 

r> 

where  tanh  0  =  -j  • 

sinh  a;  =  J  (e*  —  e"1)  =  —  sinh  (—  x)  =  —  j  sin 
cosh  x  =  %  (ex  +  e"*)  =  cosh  (—  x)  =  cos  (jz). 


tanh  x  =  6a.      ^  =  —  tanh  (—  x). 

sinh  x  +  sinh  t/  =  2  sinh  %  (x  +  y)  cosh  |  (a;  —  2/)- 
sinh  x  —  sinh  2/  =  2  cosh  %  (x  +  y)  sinh  J  (a;  —  y). 
cosh  x  +  cosh  j/  =  2  cosh  J  (a:  +  y)  cosh  |(a;  —  y) 
cosh  a;  —  cosh  y  =  2  sinh  J  (a;  +  y)  suih  J  (a;  —  y). 
cosh  x  +  sinh  x  =  ex. 
cosh  x  —  sinh  a;  =  e~x. 

x2      x* 
cosh  a?  «=  1  +  jrt  4-  7»  H"  •  •  *  • 


CALCULATION  OF  ALTERNATING  CVRRENT  PROBLEMS    267 


cosh  (x  +  jy)  =  cosh  x  cos  y  +  j  sinh  x  sin  y. 
sinh  (x  +  jy)  =  sinh  z  cos  y  +  j  cosh  z  sin  y. 
cosh  (x  dh  jy)  =  cos  (?/  =F  jx). 
sinh  (x  d=  j?/)  =  =b  j  sin  (?/  =F  jx). 
sin  (a;  db  jy)  =  ±  sinh  (y  =F  jx). 
cos  x  it 


COMPLEX   QUANTITIES 

Any  function  of  a  complex  quantity  can  be  put  in  the  fol 
lowing  form: 

f(a+jb)  =A+jB. 
Addition : 

(a+jb)  +  (c+jd)  =A+jB, 

A  =  a  +  c,  B  =  b+d. 
Subtraction : 

(a+jb)-(c+jd)=A+jB, 

A  =  a  —  c,    B  =  b  —  d. 
Multiplication: 

(a+jb)(c+jd)  =A+jB, 

A  =  ac  -  bd}    B  =  b*  +  ad. 
Division : 


.  ac  -j-  bd    j-.     be  —  ad 

Square  root: 

Va+jb  =  A+jB, 

•»     B  = 


Logarithm : 

log(a+jb)  =A+jB, 

A  =  J  log  (a2  +  62),     B  =  tan 


a 


268  FORMULAE  AND  TABLES  FOR  THE 

Exponential : 


jd  =  A  +  JB, 
A  =  aec  cos  d,     B  =  at0  sin  d. 


A  =  aec  cos  d,     B  =  —  aec  sin  d, 


=  £   +JB, 

A  =  pq€a+c  cos  (6  +  d), 
B  =  pqea+c  sin  (6  +  d). 


E4«-«  cos  (6  -  d), 


B  = 


.  _  aec  —  bfc  +  afd  +  e&6?       R  _  acf  +  efrc  —  ae  d  + 
c2  +  d2  c2  +  rf2 

^4.  =  a  +  j'6  =  r  (cos  /3  +  j  sin  /3) 


where  r  =  Va2  +  62,      tan  /?  =  -  - 

(a+jb)n  =  rnScos/3-f.7sin/3)nj: 

i  IP 

\/a-\-jb  =  rnen. 

Roots  of  unity: 


, 

^i  =  +1,  -i,    +j,    -j 

,    •  • 

=+i,    -i,    +,,    -A 


cyr. 
w/'T  ^  7T/i/  .      ^  TT/v  /7  rv    •«     r»  -i  \ 

VI  =  cos  --  hjsm  -  =e   n   (A;  =  0,1,2,   .  .  .  n  —  1), 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    269 

MISCELLANEOUS  FORMULAE 
(a  +  b)n  =  a»  +  nar-*b  +  H  (n~  1}  a"~262  +  •  •  • 

n\an~kbk  r,2    .    21 

- 


[x*  <  I]. 
[x*  <  1]. 


1         1-1    ,      1-1-3    ,      1-1-3-5    . 
--—-*--- 


a  ± ,)-,  =  i  T 


[x2  <  1]. 

1     _ 1^2    2      1-2-5   3  _    1-2-5-8     4 

[x2  <  1]. 


[x2  <  1}. 


~ 


MISCELLANEOUS  FUNCTIONS 

•j.4  .      /v.8 


22  •  4222  -  42  •  62  •  8222  •  42  •  62  •  82  •  102 


X 


10 


ber'(x) 


2!      22-42-62      22-42-62-82-102 
d  ber  (x)  _         x3  x7 

=  2~  22-42-6  +  22-42-62-82-10 


270  FORMULAE  AND  TABLES  FOR  THE 

For  x  >  6  we  have  the  following  approximate  expressions. 

,      f  x       €a  cos  j8 
ber  (x)  =      ,        > 
V2irx 

,   .  /  v       6a  sin  j8 

bei  (x)  =      ,        ) 

where 

a  =  0.707105  x  +  0.0884  x~l  -  0.046  x~s, 

j8  =  0.707105  a;  -  0.39270  -  0.0884  x-1  -  0.0625  or2  -  0.046a;-3. 

INTERPOLATION  FORMULA 


where 


CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    271 


TABLE  XXI 

TABLE  OF  NAPIERIAN  LOGARITHMS  TO  NINE  DECIMAL  PLACES  FOR  NUMBERS 

FROM  1  TO  100 


1 

0.000000000 

36 

3.583518938 

71 

4.262679877 

2 

0.693147181 

37 

3.610917913 

72 

4.276666119 

3 

1.098612289 

38 

3.637586160 

73 

4.290459441 

4 

1.386294361 

39 

3.663561646 

74 

4.304065093 

5 

1.609437912 

40 

3.688879454 

75 

4.317488114 

6 

1.791759469 

41 

3.713572067 

76 

4.330733340 

7 

1.945910149 

42 

3.737669618 

77 

4.343805422 

8 

2.079441542 

43 

3.761200116 

78 

4.356708827 

9 

2.197224577 

44 

3.784189634 

79 

4.369447852 

10 

2.302585093 

45 

3.806662490 

80 

4.382026635 

11 

2.397895273 

46 

3.828641396 

81 

4.394339155 

12 

2.484906650 

47 

3.850147602 

82 

4.406719247 

13 

2.564949357 

48 

3.871201011 

83 

4.418840608 

14 

2.639057330 

49 

3.891820298 

84 

4.430816799 

15 

2.708050201 

50 

3.912023005 

85 

4.442651256 

16 

2.772588722 

51 

3.931825633 

86 

4.454347296 

17 

2.833213344 

52 

3.951243719 

87 

4.465908119 

18 

2.890371758 

53 

3.970291914 

88 

4.477336814 

19 

2.944438979 

54 

3.988984047 

89 

4.488636370 

20 

2.995732274 

55 

4.007333185 

90 

4.499809670 

21 

3.044522438 

56 

4.025351691 

91 

4.510859507 

22 

3.091042453 

57 

4.043051268 

92 

4.521788577 

23 

3.135494216 

58 

4.060443011 

93 

4.532599493 

24 

3.178053830 

59 

4.077537444 

94 

4.543294782 

25 

3.218875825 

60 

4.094344562 

95 

4.553876892 

26 

3.258096538 

61 

4.110873864 

96 

4.564348191 

27 

3.295836866 

62 

4.127134385 

97 

4.574710979 

28 

3.332204510 

63 

4.143134726 

98 

4.584967479 

29 

3.367295830 

64 

4.158883083 

99 

4.595119850 

30 

3.401197382 

65 

4.174387270 

100 

4.605170186 

31 

3.433987204 

66 

4.189654742 

32 

3.465735903 

67 

4.204692619 

33 

3.496507561 

68 

4.219507705 

34 

3.526360525 

69 

4.234106505 

35 

3.555348061 

70 

4.248495242 

272 


FORMULAE  AND  TABLES  FOR  THE 


TABLE  XXII 
EXPONENTIAL  AND  HYPERBOLIC  FUNCTIONS 


X 

e* 

«-* 

Coshz 

Sinhx 

X 

e* 

IT" 

Coshx 

Sinhz 

0.00 

1.00000 

1.00000 

1.00000 

0.00000 

0.50 

1.64872 

0.60653 

1.12763 

0.52110 

0.01 

1.01005 

0.99005 

1.00005 

0.01000 

0.51 

1.66529 

0.60050 

1.13289 

0.53240 

0.02 

1.02021 

0.98020 

1.00020 

0.02000 

0.52 

1.68203 

0.59452 

1.13827 

0.54375 

0.03 

1.03045 

0.97045 

1.00045 

0.03000 

0.53 

1.69893 

0.58860 

1.14377 

0.55516 

0.04 

1.04081 

0.96080 

1.00080 

0.04001 

0.54 

1.71601 

0.58275 

1.14938 

0.56663 

0.05 

1.05127 

0.95123 

1.00125 

0.05002 

0.55 

1.73325 

0.57695 

1.15510 

0.57815 

0.06 

1.06184 

0.94177 

1.00180 

0.06004 

0.56 

1.75067 

0.57121 

1.16094 

0.58973 

0.07 

1.07251 

0.93239 

1.00245 

0.07006 

0.57 

1.76827 

0.56553 

1.16690 

0.60137 

0.08 

1.08329 

0.92312 

1.00320 

0.08009 

0.58 

1.78604 

0.55990 

1.17297 

0.61307 

0.09 

1.09417 

0.91393 

1.00405 

0.09012 

0.59 

1.80399 

0.55433 

1.17916 

0.62483 

0.10 

1  .  10517 

0.90484 

1.00500 

0.10017 

0.60 

1.82212 

0.54881 

1.18547 

0.63665 

0.11 

1.11628 

0.89583 

1.00606 

0.11022 

0.61 

1.84043 

0.54335 

1.19189 

0.64854 

0.12 

1  .  12750 

0.88692 

1.00721 

0.12029 

0.62 

1.85893 

0.53794 

1.19844 

0.66049 

0.13 

1.13883 

0.87810 

1.00846 

0.13037 

0.63 

1.87761 

0.53259 

1.20510 

0.67251 

0.14 

1.15027 

0.86936 

1.00982 

0.14046 

0.64 

1.89648 

0.52729 

1.21189 

0.68459 

0.15 

1.16183 

0.86071 

1.01127 

0.15056 

0.65 

1.91554 

0.52205 

1.21879 

0.69675 

0.16 

1.17351 

0.85214 

1.01283 

0.16068 

0.66 

1.93479 

0.51685 

1.22582 

0.70897 

0.17 

1.18530 

0.84366 

1.01448 

0.17082 

0.67 

1.95424 

0.51171 

1.23297 

0.72126 

0.18 

1.19722 

0.83527 

1.01624 

0.18097 

0.68 

1.97388 

0.50662 

1.24025 

0.73363 

0.19 

1.20925 

0.82696 

1.01810 

0.19115 

0.69 

1.99372 

0.50158 

1.24765 

0.74607 

0.20 

1.22140 

0.81873 

1.02007 

0.20134 

0.70 

2.01375 

0.49659 

1.25517 

0.75858 

0.21 

1.23368 

0.81058 

1.02213 

0.21555 

0.71 

2.03399 

0.49164 

1.26282 

0.77117 

0.22 

1.24608 

0.80252 

1.02430 

0.22178 

0.72 

2.05443 

0.48675 

1.27059 

0.78384 

0.23 

1.25860 

0.79453 

1.02657 

0.23203 

0.73 

2.07508 

0.48191 

1.27850 

0.79659 

0.24 

1.27125 

0.78663 

1.02894 

0.24231 

0.74 

2.09594 

0.47711 

1.28652 

0.80941 

0.25 

1.28403 

0.77880 

1.03141 

0.25261 

0.75 

2.11700 

0.47237 

1.29468 

0.82232 

0.26 

1.29693 

0.77105 

1.03399 

0.26294 

0.76 

2.13828 

0.46767 

1.30297 

0.83530 

0.27 

1.30996 

0.76338 

1.03667 

0.27329 

0.77 

2.15977 

0.46301 

1.31139 

0.84838 

0.28 

1.32313 

0.75578 

1.03946 

0.28367 

0.78 

2.18147 

0.45841 

1.31994 

0.86153 

0.29 

1.33643 

0.74826 

1.04235 

0.29408 

0.79 

2.20340 

0.45384 

1.32862 

0.87478 

0.30 

1.34986 

0.74082 

1.04534 

0.30452 

0.80 

2.22554 

0.44933 

1.33743 

0.88811 

0.31 

1.36343 

0.73345 

1.04844 

0.31499 

.  0.81 

2.24791 

0.44486 

1.34638 

0.90152 

0.32 

1.37713 

0.72615 

1.05164 

0.32549 

0.82 

2.27050 

0.44043 

1.35547 

0.91503 

0.33 

1.39097 

0.71892 

1.05495 

0.33602 

0.83 

2.29332 

0.43605 

1.36468 

0.92863 

0.34 

1.40495 

0.71177 

1.05836 

0.34659 

0.84 

2.31637 

0.43171 

1.37404 

0.94233 

0.35 

1.41907 

0.70469 

1.06188 

0.35719 

0.85 

2.33965 

0.42741 

1.38353 

0.95612 

0.36 

1.43333 

0.69768 

1.06550 

0.36783 

0.86 

2.36316 

0.42316 

1.39316 

0.97000 

0.37 

1.44773 

0.69073 

1.06923 

0.37850 

0.87 

2.38691 

0.41895 

1.40293 

0.98398 

0.38 

1.46228 

0.68386 

1.07307 

0.38921 

0.88 

2.41090 

0.41478 

1.41284 

0.99806 

0.39 

1.47698 

0.67706 

1.07702 

0.39996 

0.89 

2.43513 

0.41066 

1.42289 

1.01224 

0.40 

1.49182 

0.67032 

1.08107 

0.41075 

0.90 

2.45960 

0.40657 

1.43309 

1.02652 

0.41 

1.50682 

0.66365 

1.08523 

0.42158 

0.91 

2.48432 

0.40252 

1.44342 

1.0409C 

0.42 

1.52196 

0.65705 

1.08950 

0.43246 

0.92 

2.50929 

0.39852 

1.45390 

1.05539 

0.43 

1.53726 

0.65051 

1.09388 

0.44337 

0.93 

2.53451 

0.39455 

1.46453 

1.06998 

0.44 

1.55271 

0.64404 

1.09837 

0.45434 

0.94 

2.55998 

0.39063 

1.47530 

1.08468 

0.45 

1.56831 

0.63763 

1  .  10297 

0.46534 

0.95 

2.58571 

0.38674 

1.48623 

1.09948 

0.46 

1.58407 

0.63128 

1  .  10768 

0.47640 

0.96 

2.61170 

0.38289 

1.49729 

1.11440 

0.47 

1.59999 

0.62500 

1.11250 

0.48750 

0.97 

2.63794 

0.37908 

1.50851 

1.12943 

0.48 

1.61607 

0.61878 

1.11743 

0.49865 

0.98 

2.66446 

0.37531 

1.51988 

1.14457 

0.49 

1.63232 

0.61263 

1.12247 

0.50985 

0.99 

2.69123 

0.37158 

1.53141 

1  .  15983 

0.50 

1.64872 

0.60653 

1  .  12763 

0.52120 

1.00 

2.71828 

0.36788 

1.54308 

1  .  17520 

CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    273 


TABLE  XXII  —  (Continued} 
EXPONENTIAL  AND  HYPERBOLIC  FUNCTIONS 


2 

ex 

e~x 

Coshz 

Sinh  x 

X 

ex 

c~x 

Cosh* 

Sinhx 

1.00 

2.71828 

0.36788 

1.54308 

1.17520 

1.50 

4.48169 

0.22313 

2.35241 

2.12928 

.01 

2.74560 

0.36422 

1.55491 

1.19069 

1.51 

4.52673 

0.22091 

2.37382 

2.15291 

.02 

2.77319 

0.36059 

1.56689 

1.20630 

1.52 

4.57223 

0.21871 

2.39547 

2.17676 

.03 

2.80107 

0.35701 

1.57904 

1.22203 

1.53 

4.61818 

0.21654 

2.41736 

2:20082 

.04 

2.82922 

0.35345 

1.59134 

1.23788 

1.54 

4.66459 

0.21438 

2.43949 

2.22510 

.05 

2.85765 

0.34994 

1.60379 

1.25386 

1.55 

4.71147 

0.21225 

2.46186 

2.24961 

.06 

2.88637 

0.34646 

1.61641 

1.26996 

1.56 

4.75882 

0.21014 

2.48448 

2.27434 

.07 

2.91538 

0.34301 

1.62919 

1.28619 

1.57 

4.80655 

0.20805 

2.50735 

2.29930 

.08 

2.94468 

0.33960 

1.64214 

1.30254 

1.58 

4.85496 

0.20598 

2.53047 

2.32449 

1.09 

2.97427 

0.33622 

1.65525 

1.31903 

1.59 

4.90375 

0.20393 

2.55384 

2.34991 

1.10 

3.00417 

0.33287 

1.66852 

1.33565 

1.60 

4.95303 

0.20190 

2.57746 

2.37557 

1.11 

3.03436 

0.32956 

1.68196 

1.35240 

1.61 

5.00281 

0.19989 

2.60135 

2.40146 

1.12 

3.06485 

0.32628 

1.69557 

1.36929 

1.62 

5.05309 

0.19790 

2.62549 

2.42760 

1.13 

3.09566 

0.32303 

1.70934 

1.38631 

1.63 

5.10387 

0.19593 

2.64990 

2.45397 

1.14 

3.12677 

0.31982 

1.72329 

1.40347 

1.64 

5.15517 

0.19398 

2.67457 

2.48059 

1.15 

3.15819 

0.31664 

1.73741 

1.42078 

1.65 

5.20698 

0.19205 

2.69951 

2.50746 

1.16 

3.18993 

0.31349 

1.75171 

1.43822 

1.66 

5.25931 

0.19014 

2.72472 

2.53459 

1.17 

3.22199 

0.31037 

1.76618 

1.45581 

1.67 

5.31217 

0.18825 

2.75021 

2.56196 

1.18 

3.25437 

0.30728 

1.78083 

1.47355 

1.68 

5.36556 

0.18637 

2.77596 

2.58959 

1.19 

3.28708 

0.30422 

1.79565 

1.49143 

1.69 

5.41948 

0.18452 

2.80200 

2.61748 

1.20 

3.32012 

0.30119 

1.81066 

1.50946 

1.70 

5.47395 

0.18268 

2.82832 

2.64563 

1.21 

3.35348 

0.29820 

1.82584 

1.52764 

1.71 

5.52896 

0.18087 

2.85491 

2.67405 

1.22 

3.38719 

0.29523 

1.84121 

1.54598 

1.72 

5.58453 

0.17907 

2.88180 

2.70273 

1.23 

3.42123 

0.29229 

1.85676 

1.56447 

1.73 

5.64065 

0.17728 

2.90897 

2.73168 

1.24 

3.45561 

0.28938 

1.87250 

1.58311 

1.74 

5.69734 

0.17552 

2.93643 

2.76091 

1.25 

3.49034 

0.28650 

1.88842 

1.60192 

1.75 

5.75460 

0.17377 

2.96419 

2.79041 

1.26 

3.52542 

0.28365 

1.90454 

1.62088 

1.76 

5.81244 

0.17204 

2.99224 

2.82020 

.27 

3.56085 

0.28083 

1.92084 

1.64001 

1.77 

5.87085 

0.17033 

3.02059 

2.85026 

.28 

3.59664 

0.27804 

1.93734 

1.65930 

.78 

5.92986 

0.16864 

3.04925 

2.88061 

.29 

3.63279 

0.27527 

1.95403 

1.67876 

.79 

5.98945 

0.16696 

3.07821 

2.91125 

.30 

3.66930 

0.27253 

1.97091 

1.69838 

.80 

6.04965 

0.16530 

3.10747 

2.94217 

.31 

3.70617 

0.26982 

1.98800 

1.71818 

.81 

6.11045 

0.16365 

3.13705 

2.97340 

1.32 

3.74342 

0.26714 

2.00528 

1.73814 

.82 

6.17186 

0.16203 

3.16694 

3.00492 

1.33 

3.78104 

0.26448 

2.022Z6 

1.75828 

.83 

6.23389 

0.16041 

3.19715 

3.03674 

1.34 

3.81904 

0.26185 

2.04044 

1.77860 

.84 

6.29654 

0.15882 

3.22768 

3.06886 

1.35 

3.85743 

0.25924 

2.05833 

1.79909 

1.85 

6.35982 

0.15724 

3.25853 

3.10129 

1.36 

3.89619 

0.25666 

2.07643 

1.81977 

1.86 

6.42374 

0.15567 

3.28970 

3.13403 

1.37 

3.93535 

0.25411 

2.09473 

1.84062 

1.87 

6.48830 

0.15412 

3.32121 

3.16709 

1.38 

3.97490 

0.25158 

2.11324 

1.86166 

1.88 

6.55350 

0.15259 

3.35305 

3.20046 

1.39 

4.01485 

0.24908 

2.13196 

1.88289 

1.89 

6.61937 

0.15107 

3.38522 

3.23415 

1.40 

4.05520 

0.24660 

2.15090 

1.90430 

1.90 

6.68589 

0.14957 

3.41773 

3.26816 

1.41 

4.09596 

0.24414 

2.17005 

1.92591 

1.91 

6.75309 

0.14808 

3.45058 

3.30250 

.42 

4.13712 

0.24171 

2.18942 

1.94770 

1.92 

6.82096 

0.14661 

3.48378 

3.33718 

.43 

4.17870 

0.23931 

2.20900 

1.96970 

.93 

6.88951 

0.14515 

3.51733 

3.37218 

.44 

4.22070 

0.23693 

2.22881 

1.99188 

1.94 

6.95875 

0.14370 

3.55123 

3.40752 

.45 

4.26311 

0.23457 

2.24884 

2.01428 

.95 

7.02869 

0.14227 

3.58548 

3.44321 

.46 

4.30596 

0.23224 

2.26910 

2.03686 

.96 

7.099330.14086 

3.62009 

3.47923 

.47 

4.34924 

0.22993 

2.28958 

2.05965 

.97 

7.170680.13946 

3.65507 

3.51561 

.48 

4.39295 

0.22764 

2.31029 

2.08265 

1.98 

7.24274:0.13807 

3.69041 

3.55234 

.49 

4.43710 

0.22537 

2.33123 

2.10586 

1.99 

7.315330.13670 

3.72611 

3.58942 

1.50 

4.48169 

0.22313 

2.35241 

2.12928 

2.00 

7.389060.13534 

3.76220 

3.62686 

274 


FORMULAE  AND  TABLES  FOR  THE 


TABLE  XXII  —  (Continued) 
EXPONENTIAL  AND  HYPERBOLIC  FUNCTIONS 


X 

<f 

«-* 

Coshx 

Sinhx 

X 

ex 

e-z 

Coshx 

Sinhz 

2.0 

7.38906 

0.13534 

3.76220 

3.62686 

4.0 

54.5982 

0.01832 

27.3082 

27.2899 

2.1 

8.16617 

0.12246 

4.14431 

4.02186 

4.1 

60.3403 

0.01657 

30.1784 

30.1619 

2.2 

9.02501 

0.11080 

4.56791 

4.45711 

4.2 

66.6863 

0.01500 

33.3507 

33.3357 

2.3 

9.97418 

0.10026 

5.03722 

4.93696 

4.3 

73.69980.01357 

36.8567 

36.8431 

2.4 

11.0232 

0.09072 

5.55695 

5.46623 

4.4 

81.4509 

0.01228 

40.7316 

40.7193 

2.5 

12.1825 

0.08208 

6.13229 

6.05020 

4.5 

90.0171 

0.01111 

45.0141 

45.0030 

2.6 

13.4637 

0.07427 

5.76900 

6.69473 

4.6 

99.4843 

0.01005 

49.7472 

49.7371 

2.7 

14.8797 

0.06721 

7.47347 

7.40626 

4.7 

109.947 

0.00910 

54.9781 

54.9690 

2.8 

16.4446 

0.06081 

8.25273 

8.19192 

4.8 

121.510 

0.00823 

60.7593 

60.7511 

2.9 

18.1741 

0.05502 

9.11458 

9.05956 

4.9 

134.290 

0.00745 

67.1486 

67.1412 

3.0 

20.0855 

0.04979 

10.0677 

10.0179 

5.0 

148.413 

0.00674 

74.2099 

74.2032 

3.1 

22.1980 

0.04505 

11.1215 

11.0765 

5.1 

164.022 

0.00610 

82.0140 

82.0079 

3.2 

24.5325 

0.04076 

12.2866 

12.2459 

5.2 

181.272 

0.00552 

90.6388 

90.6333 

3.3 

27.1126 

0.03688 

13.5747 

13.5379 

5.3 

200.337 

0.00499 

100.171 

100.167 

3.4 

29.9641 

0.03337 

14.9987 

14.9654 

5.4 

221.406 

0.00452 

110.705 

110.701 

3.5 

33.1155 

0.03020 

16.5728 

16.5426 

5.5 

244.692 

0.00409 

122.348 

122.344 

3.6 

36.5982 

0.02732 

18.3128 

18.2855 

5.6 

270.426 

0.00370 

135.215 

135.211 

3.7 

40.4473 

0.02472 

20.2360 

20.2113 

5.7 

298.867 

0.00335 

149.435 

149.432 

3.8 

44.7012 

0.02237 

22.3618 

22.3394 

5.8 

330.300 

0.00303 

165.151 

165.148 

3.9 

49.4024 

0.02024 

24.7113 

24.6911 

5.9 

365.037 

0.00274 

182.520 

182.517 

4.0 

54.5982 

0.01832 

27.3082 

27.2899 

6.0 

403.429 

0.00248 

201.716 

201.713 

CALCULATION  OF  ALTERNATING  CURRENT  PROBLEMS    275 


TABLE  XXIII 
TRIGONOMETRIC  FUNCTIONS 


z 

Cosz 

Sinz 

z 

Cosz 

Sin  z 

z 

cos  z 

Sinz 

0.20 

0.9801 

0.1987 

0.82 

0.6822 

0.7311 

1.44 

0.1304 

0.9915 

0.22 

0.9759 

0.2182 

0.84 

0.6675 

0.7446 

1.46 

0.1106 

0.9939 

0.24 

0.9713 

0.2377 

0.86 

0.6524 

0.7578 

1.48 

0.0907 

0.9959 

0.26 

0.9664 

0.2571 

0.88 

0.6371 

0.7707 

1.50 

0.0707 

0.9975 

0.28 

0.9611 

0.2763 

0.90 

0.6216 

0.7833 

1.52 

0.0508 

0.9987 

0.30 

0.9553 

0.2956 

0.92 

0.6059 

0.7956 

1.54 

0.0308 

0.9995 

0.32 

0.9492 

0.3145 

0.94 

0.5898 

0.8076 

1.56 

0.0108 

0.9999 

0.34 

0.9428 

0.3335 

0.96 

0.5735 

0.8192 

1.58 

-0.0092 

0.9999 

0.36 

0.9359 

0.3522 

0.98 

0.5590 

0.8305 

1.60 

-0.0292 

0.9995 

0.38 

0.9287 

0.3709 

1.00 

0.5403 

0.8415 

1.62 

-0.0492 

0.9988 

0.40 

0.9211 

0.3894 

.02 

0.5235 

0.8521 

1.64 

-0.0692 

0.9976 

0.42 

0.9131 

0.4078 

.04 

0.5062 

0.8624 

.66 

-0.0891 

0.9960 

0.44 

0.9048 

0.4259 

.06 

0.4889 

0.8724 

.68 

-0.1090 

0.9940 

0.46 

0.8961 

0.4439 

.08 

0.4713 

0.8820 

.70 

-0.1289 

0.9917 

0.48 

0.8870 

0.4618 

.10 

0.4536 

0.8912 

.72 

-0.1486 

0.9889 

0.50 

0.8776 

0.4794 

.12 

0.4357 

0.9001 

.74 

-0.1684 

0.9857 

0.52 

0.8678 

0.4963 

.14 

0.4176 

0.9086 

.76 

-0.1881 

0.9822 

0.54 

0.8577 

0.5137 

.16 

0.3993 

0.9168 

.78 

-0.2076 

0.9782 

0.56 

0.8473 

0.5312 

.18 

0.3809 

0.9246 

.80 

-0.2272 

0.9738 

0.58 

0.8365 

0.5479 

.20 

0.3624 

0.9320 

1.82 

-0.2466 

0.9691 

0.60 

0.8253 

0.5646 

.22 

0.3436 

0.9391 

1.84 

-0.2660 

0.9640 

0.62 

0.8138 

0.5810 

.24 

0.3248 

0.9458 

1.86 

-0.2852 

0.9585 

0.64 

0.8021 

0.5972 

1.26 

0.3058 

0.9521 

1.88 

-0.3043 

0.9526 

0.66 

0.7900 

0.6131 

1.28 

0.2867 

0.9580 

1.90 

-0.3233 

0.9463 

0.68 

0.7776 

0.6288 

1.30 

0.2675 

0.9636 

1.92 

-0.3422 

0.9396 

0.70 

0.7648 

0.6442 

1.32 

0.2482 

0.9687 

1.94 

-0.3609 

0.9326 

0.72 

0.7518 

0.6594 

1.34 

0.2287 

0.9735 

1.96 

-0.3795 

0.9252 

0.74 

0.7384 

0.6743 

1.36 

0.2093 

0.9779 

1.98 

-0.3979 

0.9174 

0.76 

0.7248 

0.6889 

1.38 

0.1896 

0.9817 

2.00 

-0.4162 

0.9093 

0.78 

0.7109 

0.7033 

1.40 

0.1700 

0.9855 

0.80 

0.6967 

0.7174 

1.42 

0.1502 

0.9887 

276 


ALTERNATING  CURRENT  PROBLEMS 


TABLE  XXIV 

BINOMIAL  COEFFICIENTS  FOR  INTERPOLATION  BY  DIFFERENCES 
(For  use  in  interpolation  formula) 


1 

Coefficients  of 

t 

Coefficients  of 

t 

Coefficients  of 

t 

Coefficients  of 

A2 

A3 

A2 

A3 

A2 

A3 

A2 

A3 

.01 

-.005 

+  .003 

.26 

-.096 

+  .056 

.51 

-.125 

+  .062 

.76 

-.091 

+  .038 

.02 

-.010 

+  .006 

.27 

-.099 

+  .057 

.52 

-.125 

+  .062 

.77 

-.089 

+  .036 

.03 

-.015 

+  .010 

.28 

-.101 

+  .058 

.53 

-.125 

+  .061 

.78 

-.086 

+  .035 

.04 

-.019 

+  .013 

.29 

-.103 

+  .059 

.54 

-.124 

+  .060 

.79 

-.083 

+  .033 

.05 

-.024 

+  .015 

.30 

-.105 

+  .060 

.55 

-.124 

+  .060 

.80 

-.080 

+  .032 

.06 

-.028 

+  .018 

.31 

-.107 

+  .060 

.56 

-.124 

+  .059 

.81 

-.077 

+  .031 

.07 

-.033 

+  .021 

.32 

-.109 

+  .061 

.57 

-.123 

+  .058 

.82 

-.074 

+  .029 

.08 

-.037 

+  .024 

.33 

-.111 

+  .062 

.58 

-.122 

+  .058 

.83 

-.071 

+  .028 

.09 

-.041 

+  .026 

.34 

-.112 

+  .062 

.59 

-.121 

+  .057 

.84 

-.067 

+  .026 

.10 

-.045 

+  .028 

.35 

-.114 

+  .063 

.60 

-.120 

+  .056 

.85 

-.064 

+  .024 

.11 

-.049 

+  .031 

.36 

-.115 

+  .063 

.61 

-.119 

+  .055 

.86 

-.060 

+  .023 

.12 

-.053 

+  .033 

.37 

-.117 

+  .063 

.62 

-.118 

+  .054 

.87 

-.057 

+  .021 

.13 

-.057 

+  .035 

.38 

-.118 

+  .064 

.63 

-.117 

+  .053 

.88 

-.053 

+  .020 

.14 

-.060 

+  .037 

.39 

-.119 

+  .064 

.64 

-.115 

+  .052 

.89 

-.049 

+  .018 

.15 

-.064 

+  .039 

.40 

-.120 

+  .064 

.65 

-.114 

+  .051 

.90 

-.045 

+  .016 

.16 

-.067 

+  .041 

.41 

-.121 

+  .064 

.66 

-.112 

+  .050 

.91 

-.041 

+  .015 

.17 

-.071 

+  .043 

.42 

-.122 

+  .064 

.67 

-.111 

+  .049 

.92 

-.037 

+  .013 

.18 

-.074 

+  .045 

.43 

-.123 

+  .064 

.68 

-.109 

+  .048 

.93 

-.033 

+  .012 

.19 

-.077 

+  .047 

.44 

-.123 

+  .064 

.69 

-.107 

+  .047 

.94 

-.028 

+  .010 

.20 

-.080 

+  .048 

.45 

-.124 

+  .064 

.70 

-.105 

+  .045 

.95 

-.024 

+  .008 

.21 

-.083 

+  .049 

.46 

-.124 

+  .064 

.71 

-.103 

+  .044 

.96 

-.019 

+  .007 

.22 

-.086 

+  .051 

.47 

-.125 

+  .064 

.72 

-.101 

+  .043 

.97 

-.015 

+  .005 

.23 

-.089 

+  .052 

.48 

-.125 

+  .063 

.73 

-.099 

+  .042 

.98 

-.010 

+  .003 

.24 

-.091 

+  .053 

.49 

-.125 

+  .063 

.74 

-.096 

+  .040 

.99 

-.005 

+  .002 

.25 

-.094 

+  .055 

.50 

-.125 

+  .063 

.75 

-.094 

+  .039 

1.00 

-.000 

+  .000 

INDEX. 


Abraham,  M.,  190. 
Agnew,  P.  G.,  167,  169. 
Alexanderson,  E.  F.,  40. 
Alternating  current  circuits,  129. 
Antennae  horizontal,  capacity  of,  108. 
Attenuation    constant    of   line,    202. 

of  loaded  line,  203. 

of  non  inductive  line,  204. 
Attraction  between  circular  currents, 

51. 

Arrival  curves  on  cables,  235. 
Average  value  of  alternating  wave, 
128. 

Benischke,  160. 

Bethenod,  J.,  160. 

Berg,  E.  J.,  24. 

Blondel,  A.,  263. 

Brooks  Morgan  and  H.  M.  Turner, 


Cable,  alternating  current  resistance, 
7. 

grounded  lead  covering,   effective 
resistance    and    reactance,     23. 

capacity  of,  111. 

transatlantic,   arrival  curves,   235. 

transatlantic,    current     and    volt- 
age distribution,  232. 
Capacity,  parallel  or  series  arrange- 
ment, 87. 

of  parallel  plate  condenser,  88. 

of  concentric  spheres,  90. 

of  disk  insulated  in  free  space,  91. 

of  linear  conductors,  94. 

of  overhead  wires,  94. 

of  grounded  wires,  96. 

of  looped  conductor,  100. 

coefficients  of  two  spheres,  91. 

of  two  metallic  circuits  suspended 
on  the  same  pole,  105. 


Capacity  of  three-phase  transmission 

lines,  106. 

of  horizontal  antennae,  108. 
of  concentric  cylinders,  109. 
of  concentric  cables,  102. 
of  two  core  cable,  115. 
of  three  core  cable,  117. 
specific  inductive  of  solids,  table, 

123. 
specific  inductive  of  liquids,  table, 

125. 

distributed,  200. 

Charging     condenser,     current     and 

voltage,    logarithmic   case,    176. 

Charging     condenser,     current     and 

voltage,  critical  case,  178. 
Charging  condenser,  current  and  volt- 
age, trigonometric  case,  178. 
Circles    coaxial,    mutual   inductance 

of,    48. 
Circular  currents,  attraction  between, 

51. 

Circular  ring,  self-inductance  of,  59. 
Circuits,  alternating  current,   129. 
in     parallel,     each     branch     con- 
taining   inductance    and    resist- 
ance, 136. 

in  parallel,  each  branch  having 
inductance,  capacity  and  re- 
sistance, 142. 

in  parallel,  mutual  inductance  be- 
tween the  circuits,  143. 
containing     capacity    and    resist- 
ance, current  rise  on  closing,  174. 
containing  inductance  and  resist- 
ance,   current    rise   on    closing, 
171. 

divided,  values  of  transient  cur- 
rents, 186. 

having  mutual  inductance,  values 
of  transient  currents,  190. 


277 


278 


INDEX 


Coil,  single  layer,  self  inductance  of, 

56. 
circular,   rectangular  section,   self 

inductance  of,  58. 
solenoid,  single  layer,  self  induc- 
tance of,  56. 

solenoid,    several    layers,    self   in- 
ductance of,  58. 
Coils,  alternating  current  resistance 

of,  14. 

coaxial,  mutual  inductance  of,  52. 
Coils,  toroidal,  magnetic  flux  distri- 
bution in  iron  core  of,  39. 
toroidal,  self-inductance  of,  63. 
Coefficient  of  coupling,  195. 
Cohen,  Louis,  15,  48,  58,  69,  75,  105, 

108. 
Complex   waves,  effective  value  of, 

128. 

waves,  power  factor  of,  145. 
quantities,  formulae,  267. 
Concentric  conductors,  resistance  of, 

9. 
conductors,     hollow     inner     core, 

resistance  of,  11. 

conductors,  self  inductance  of,  71. 
cable,  capacity  of,   112. 
Condenser     charge    and     discharge, 

logarithmic  case,  176. 
charge  and  discharge,  critical  case, 

178. 
charge  and  discharge,  trigonometric 

case,  178. 

Conductance  leakage,   between   cyl- 
inder and  infinite  plane,  27. 
leakage,   concentric   cylinders,    29. 
Conductor    return,    alternating    cur- 
rent resistance,  7. 
concentric,  alternating  current  re- 
sistance, 9. 
Conductors    cylindrical,    alternating 

current  resistance,  2. 
cylindrical,    high-frequency   resist- 
ance, 3. 

cylindrical,  current  penetration,  6. 
flat,     alternating     current     resist- 
ance, 12. 
flat,  current  penetration,  13. 


Conductors  slot  wound,  alternating 
current  resistance,  18. 

enclosed  in  iron  pipes,  eddy  current 
losses,  20. 

linear,  leakage  conductance,  27. 

linear,  inductance  of,  64. 

linear,  mutual  inductance  of,  68. 

split,  inductance  of,  69. 

concentric,  inductance  of,  71. 

compound,  inductance  of,  73. 

linear,  capacity  of,  94. 
Coupling  coefficient,  inductive  195. 

direct,  197. 

Coupling  for  maximum  current  in 
secondary  of  a  resonance  trans- 
former, 161. 

Cramp  W.  and  C.  F.  Smith,  165. 
Current    penetration    in    flat    con- 
ductors, 13. 

penetration     in     cylindrical     con- 
ductors, 6. 

maximum    in    secondary  of  reso- 
nance transformer,  161. 

rise  in  inductive  circuit,  172. 

decay  in  inductive  circuit,  173. 

rise    on     closing     an    alternating 
current  inductive  circuit,  174. 

rise  on  closing  a  circuit  containing 
resistance  and  capacity,    174. 

effective  value  of  oscillatory  dis- 
charge, 181. 

and  voltage  propagation  on  long 
lines,  general  equations,  201. 

distribution  on  long  line,  general 
solution,  212. 

rise  on  cables,  232. 
Curve  factor,  129. 

Cylinders,  concentric,  leakage  con- 
ductance, 29. 

concentric,  self-inductance  of,  71. 

concentric,  capacity  of,  109. 

iron,    magnetic    flux    distribution, 
35. 

concentric,  resistance  of,  9. 

Damping     factor,     oscillatory     dis- 
charge, 178. 
Decrement,  logarithmic,  180. 


INDEX 


279 


Direct  coupling,  197. 

Distributed  inductance  and  capacity, 

200. 
Divided  circuits,  transient  terms  on 

changing    resistance    in    either 

branch,  186. 
Drude,  P.,  194. 

Eddy  current,  energy  losses  in  iron 

plates,  30. 
current    energy    losses     in     iron 

cylinders,  36. 

current,    energy   loss    in  an   iron 
pipe  enclosing  conductor  carry- 
ing on  alternating  current,  20. 
Effective  resistance  and  reactance  of 
grounded  lead  covered  cable,  24. 
Effective  value    of    current,    oscilla- 
tory  discharge,  181. 
Efficiency  iron  core  transformer,  166. 

in  power  transmission,  147. 
Empirical  formula  for  the  calculation 
of  the  self-inductance  of  coils, 
59. 

Equations,  general,  for  current  and 
voltage  propagation  on  wires, 
201. 

Essau,  A.,  18. 

Exponential    and    hyperbolic    func- 
tions, table,  272. 
and  logarithmic  formulae,  264. 

Field,  A.  B.,  18. 
Field,  M.  B.,  22. 
Flat  conductors,  alternating  current 

resistance  of,  13. 
conductors,    current    distribution, 

12. 

Fleming,  J.  A.,  2,  88,  175,  194. 
Foppl,  A.,  190. 

Form  factor,  current  wave,  129. 
Formulae  in  complex  quantities  for 
power  transmission  calculations, 
220. 

exponential  and  logarithmic,  264. 
hyperbolic,  266. 
mathematical,  264. 
trigonometric,  265. 


Formulae  for  complex  quantities,  267. 

interpolation,  270. 

miscellaneous,  269. 
Functions,  miscellaneous,  269. 
Fowle,  F.  F.,  74,  94. 
Frequency   of   oscillatory   discharge, 

179. 

Frequencies,  oscillation  transformer, 
196. 

direct  coupled  circuits,  198. 

Gray,  Andrew,  133. 

Heaviside,  O.,  68,  94. 
Hyperbolic  formulae,  266. 
Hyperbolic  functions,  table  of,  243. 

Impedances  in  series,  133. 
Inductance   and   resistance   shunted 
by  a  condenser,   current  distri- 
bution, 137. 

mutual,  two  coaxial  circles,  48. 

mutual,  two  coaxial  coils,  52. 
Inductance  mutual,  concentric  coax- 
ial solenoids,  54. 

mutual,  rectangles,  62. 

self,  single  layer  coil,  56. 

self,    correction    for    thickness    of 
insulation  on  windings,  57. 

self,    circular    coil    of    rectangular 
section,  58. 

self,    solenoid    of    several    layers, 
58. 

self,  circular  ring,  59. 

self,  rectangle,  62. 

self,  toroidal  coil,  63. 

self,  empirical  formula,  59. 

linear  conductors,  64. 

three-phase  transmission  lines,  66. 

grounded  conductors,  68. 

of  two  parallel  hollow  cylindrical 
conductors,  70. 

concentric  conductors,  71. 

series  or  parallel  arrangement,  75. 

of  wires  in  parallel,  76. 

table    of   values    giving    end    cor- 
rection of  inductance  of  coils,  80. 

effective,  transformer,  160. 


280 


INDEX 


Inductive  load  shunted  by  condenser, 

139. 

Inductive  circuit,  current  rise,  172. 
Induction,  distribution  in  iron  pipe 

enclosing  an  alternating  current, 

20. 

magnetic  flux  in  iron  plates,  31. 
magnetic  flux  in  iron  cylinders,  36. 
Interpolation  formula,  270. 
Iron,  magnetic  flux  distribution  and 

eddy  current  losses,  30. 

Karapetoff,  V.,  107,  165. 

Kennelly,  A.  E.,  28,  95,  100,  103,  2"63. 

Kirchoff,  G.,  88. 

LaCour,  J.  L.,  and  O.  S.  Bragstad, 

7,  68,  128,  144,  187. 
Leakage  conductance,  27. 
Lichtenstein,  L.,  111. 
Lines   transmission,   transient   terms 
on     short     circuiting     load     or 
generator,  188. 
Line    infinite,    current    and    voltage 

distribution,  213. 

finite  length  open  at  receiving  end, 
current  and  voltage  distribution, 
214. 

grounded   at   receiving   end,    217. 
short   circuited   at  receiving  end, 

218. 

oscillations,  236. 

Load,    inductive    shunted    by    con- 
denser, 139. 
Logarithmic    charge    and    discharge, 

176. 

decrement,  180. 

current  rise  on   closing   an  alter- 
nating current  circuit,  185. 
formulae,  264. 
Logarithms,  naperian,  table  of,  271. 

Magnetic   flux    distribution   in    iron 

plates,  31. 
flux  distribution  in  iron  cylinders, 

35. 
flux   distribution   in   iron   core   of 

toroidal  coil,  39. 


Mathematical  formulae,  264. 
Maxwell,  C.,  91,  159. 
Mie,  G.,  8. 

Miller,  W.  E.,  223,  243. 
Miscellaneous  formulae,  269. 
functions,  269. 

Nogaoka,  H.,  51,  56. 
Nicholson,  J.  W.,  17. 

Oberleck,  A.,  194. 
Oscillatory  discharge,  178. 

discharge,   effective  value  of  cur- 
rent, 181. 

current  rise  on   closing   an   alter- 
nating current  circuit,  183. 
Oscillation  transformer,  194. 
Oscillations  on  line  grounded  at  one 
end  and  open  at  the  other  end, 
237. 
on   line   grounded   at    both   ends, 

239. 

on  line  open  at  both  ends,  240. 
on  line  closed  upon  itself,  241. 
Pedersen,  P.  O.,  4,  65. 
Pender,  H.,  and  H.  S.  Osborne,  100. 
Perrine,  F.  A.  C.,  and  F.  G.  Baum, 

106,  150. 

Pierce,  G.  W.,160. 
Power  factor,  144. 

factor,  complex  wave,  145. 
maximum  supply,  147. 
transmission,  efficiency,  147. 
transmission,  inductive  load,   148. 
transmission,     line     capacity     ne- 
glected, 151. 

transmission,  line  capacity  shunted 

across  the  middle  of  the  line,  153. 

transmission,     half    line    capacity 

shunted  across  each  end  of  line, 

155. 

transmission    calculations,    hyper- 
bolic formulae,  222. 
transmission  calculations,  approxi- 
mate formulae,  222. 
transmission,  generator  voltage  and 
impedance     at     receiving     end 
given,  226. 


INDEX 


281 


Power  transmission,  formulae  ex- 
pressed in  complex  quantities, 
220. 

Pupin,  M.  I.,  263. 

Rayleigh,  Lord,  3,  143. 

Reactance    effective,    grounded    lead 

covered  cable,  23. 

Resistance,  cylindrical  conductors,  2. 
high-frequency,     cylindrical     con- 
ductors, 3. 
general     formula     for     cylindrical 

conductors,  4. 

curves   showing   change   in   resist- 
ance of  copper  wire  for  alternat- 
ing currents,  5. 
return  conductor,  7. 
concentric  conductor,  9. 
alternating    current    of    flat    con- 
ductors, 12. 

alternating  current  of  coils,  14. 
slot  wound  conductors,  18. 
effective  transformer,  160. 
effective    grounded    lead    covered 

cable,  23. 
Resistance  effective,  ungrounded  lead 

covered  cable,  25. 
increase  due  to  eddy  currents  in 

iron  core  of  toroidal  coil,  40. 
increase  due  to  iron  core  in  cylin- 
drical coil,  36. 

specific,  of  metallic  wires,  47. 
Regulation  iron  core  transformer,  167. 
Resonance     in     alternating     current 

circuits,  129. 
transformer,  160,  162. 
Roessler,  G.,  263. 

Root,  mean  square  value  of  alternat- 
ing wave,  128. 

Rosa,  E.  B.,  and  Louis  Cohen,  64. 
Russell,  A.,  31,  63,  70,  91,  111,  133, 
159. 

Soibt,  G.,  197. 

Solenoids,     coaxial,     mutual    induc- 
tance of,  54. 
self  inductance  of,  56. 


Sommerfeld,  A.,  16. 

Spheres  concentric,   capacity  of,   90. 

capacity  coefficients,  91. 
Specific  inductive  capacity  of  solids, 
123. 

inductive  capacity  of  liquids,  125. 

resistance  of  metallic  wires,  7. 
Steinmetz,   C.   P.,    12,   31,    137,   150, 
165,  185,  190,  263.  • 

Stone,  J.  S.,  100. 

Thomas,  P.  H.,  150. 
Thomson,  J.  J.,  6,  143. 
Transformer,  air  core,  159. 
resonance,  160. 
condenser     in     secondary     circuit, 

161. 

condenser  in  primary  circuit,  163. 
iron  core,  general  design  formula, 

165. 

the  current,  169. 
effective  resistance  of,  160. 
effective  reactance  of,  160. 
condition  for  resonance,  162. 
Transformer,  efficiency  formula,  166. 
regulation  formula,  167. 
oscillation,  194. 

Transformation  ratio,  potential  trans- 
former, 166. 

ratio,  current  transformer,  169. 
Transmission  lines,  transient  current 
and  voltage  on  short  circuiting 
load  or  generator,  188. 
of  power,  see  power  transmission, 
lines,  see  inductance  and  capacity 

of  lines. 
Trigonometric  formulae,  265. 

functions,  table  of,  275. 
Turner,    H.    M.,   see    Brooks,    Mor- 
gan. 

Vaschy,  A.,  74. 

Velocity  of  propagation  of  elec- 
tromagnetic waves  on  wires, 
202. 

of    propagation    on    loaded    lines, 
203. 


282 


INDEX 


Velocity  of  propagation  on  noninduc- 

tive  lines,  204. 

Voltage  distribution  on  lines,  general 
solution,  212. 

rise  on  cables,  232. 

rise  on  transmission  line  on  short- 
circuiting  load,  189. 

transformation  in  iron  core  trans- 
former, 196. 

transformation  in  oscillation  trans- 
former, 169. 


Wien,  M.,  15. 

Wire,   copper,   depth    of    alternating 

current  penetration,  7. 
copper,     alternating     current     re- 
sistance curves,  5. 
aluminum,    depth    of    alternating 

current  penetration,  7. 
iron,  depth  of  alternating  current 

penetration,  7. 

Wave     length     constant     of     line, 
202. 


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